[编程题]推倒吧骨牌
思路，从这往右遍历，用flag记录之前是否有右倾；用pos记录上次右倾位置；遍历无非两种情况：
当前是左倾：1.在最左侧，直接往左边倒完。2.前面存在右倾：相拥而倒；
当前是右倾：1.前面有右倾：处理前面的右倾，更新pos;2:前面无右倾，设置flag和pos；
遍历结束后再处理是否有最后残留的那个flag；（码量很少，我只是大括号打得多）
#include<iostream>
#include<string>
using namespace std;
int main()
{
string data;
while(cin>>data)
{
int rflag=0;
int pos;
int len=data.size();
for(int i=0;i<len;i++)
{
if(data[i]=='L')
{
int cur=i-1;
if(rflag==0)
{
while(cur>=0&&data[cur]=='.')
{
data[cur]='L';cur--;
}
}
else
{
int l=pos+1;
while(cur>l)
{
data[cur]='L';
data[l]='R';
l++;cur--;
}
rflag=0;
}
}
if(data[i]=='R')
{
if(rflag==0)
{
rflag=1;
pos=i;
}
else{
pos=i;
int cur=i-1;
while(cur>=0&&data[cur]=='.')
{
data[cur]='R';cur--;
}
}
}
}
if(rflag)
{
int cur=pos+1;
while(cur<len)
{
data[cur]='R';
cur++;
}
}
cout<<data<<endl;
}
return 0;
}
#include<string>
using namespace std;
int main()
{
string data;
while(cin>>data)
{
int rflag=0;
int pos;
int len=data.size();
for(int i=0;i<len;i++)
{
if(data[i]=='L')
{
int cur=i-1;
if(rflag==0)
{
while(cur>=0&&data[cur]=='.')
{
data[cur]='L';cur--;
}
}
else
{
int l=pos+1;
while(cur>l)
{
data[cur]='L';
data[l]='R';
l++;cur--;
}
rflag=0;
}
}
if(data[i]=='R')
{
if(rflag==0)
{
rflag=1;
pos=i;
}
else{
pos=i;
int cur=i-1;
while(cur>=0&&data[cur]=='.')
{
data[cur]='R';cur--;
}
}
}
}
if(rflag)
{
int cur=pos+1;
while(cur<len)
{
data[cur]='R';
cur++;
}
}
cout<<data<<endl;
}
return 0;
}
编辑于 2019-08-23 14:59:27
回复(2)
""" LR匹配问题，记录之前方向，分4类讨论LL/RL/RR/LR
"""
import math
if __name__ == "__main__":
s = list(input().strip())
ans = s[:]
t = pl = pr = 0
flag = '0'
while t < len(s):
while t < len(s) and s[t] == '.': t += 1
if t >= len(s): break
if s[t] == 'L':
if flag == '0' or flag == 'L':
for i in range(pl, t):
ans[i] = 'L'
else:
pl = t
for i in range(pr, math.ceil((pr + pl) / 2)):
ans[i] = 'R'
for i in range(math.floor((pr + pl) / 2) + 1, pl):
ans[i] = 'L'
flag = 'L'
elif s[t] == 'R':
if flag == 'R':
for i in range(pr + 1, t):
ans[i] = 'R'
pr = t
flag = 'R'
t += 1
if pr > pl:
for i in range(pr, len(s)):
ans[i] = 'R'
print(''.join(ans))
编辑于 2019-07-11 10:08:30
回复(0)
s = input()
stat = [(i, char) for i, char in enumerate(s) if char != '.']
stat = [(-1, 'L')] + stat + [(len(s), 'R')]
#print()
#print(s)
for (i,ci), (j, cj) in zip(stat[:-1], stat[1:]):
#print(i, j, ci, cj)
if ci == cj:
if i < 0:
s = ci * (j-i) + s[j+1:]
elif j == len(s):
s = s[:i] + ci * (j-i) + s[j+1:]
else:
s = s[:i] + ci * (j-i+1) + s[j+1:]
#print(s)
elif ci > cj: # R > L
half = (j-i+1) // 2
rest = (j-i+1) % 2
s = s[:i] + 'R'*half + '.'*rest + 'L'*half + s[j+1:]
#print(i, j, s)
#print(s, i, j, s[:i], s[j+1:])
#print('LL.RR.LLRRRLLL.')
print(s)
参考LEETCODE838 https://leetcode.com/articles/push-dominoes/
发表于 2019-02-28 13:45:54
回复(1)
常规做法，注意coding细节即可
1. import java.util.Scanner;
2. import static java.lang.System.in;
3. public class Main {
4. public static void main(String[] args) {
5. Scanner sc = new Scanner(in);
6. String str = sc.nextLine();
7. char[] arr = ("L" + str + "R").toCharArray(); //首尾加一个字符方便处理
8. int begin = 0;
9. for (int i = 1; i < arr.length; i++) {
10. if (arr[i] != '.') {
11. trans(arr, begin, i);
12. begin = i;
13. }
14. }
15. StringBuilder sb = new StringBuilder();
16. for (int i = 1; i < arr.length - 1; i++) {
17. sb.append(arr[i]);
18. }
19. System.out.println(sb.toString());
20. }
21. public static void trans(char[] arr, int begin, int end) {
22.
23. for (int i = begin + 1, j = end - 1; i <= j; i++, j--) {
24. if (arr[begin] == 'L' && arr[end] == 'L') {
25. arr[i] = 'L';
26. arr[j] = 'L';
27. } else if (arr[begin] == 'R' && arr[end] == 'R') {
28. arr[i] = 'R';
29. arr[j] = 'R';
30. } else if (arr[begin] == 'R' && arr[end] == 'L' && i != j) {
31. arr[i] = 'R';
32. arr[j] = 'L';
33. }
34. }
35. }
36. }
发表于 2019-08-02 22:11:14
回复(0)
#include <bits/stdc++.h>
using namespace std;
int main(){
string s;
cin>>s;
int l = s.length();
int L[l],R[l];
for(int i=0, x=l;i<l;i++){
if(s[i]=='R'){
x = l;
R[i] = x--;
}else if(s[i]=='.' && x<l)
R[i] = x--;
else{
R[i] = 0;
x = l;
}
}
for(int i=l-1,x=l;i>=0;i--){
if(s[i]=='L'){
x = l;
L[i] = x--;
}else if(s[i]=='.' && x<l)
L[i] = x--;
else{
L[i] = 0;
x = l;
}
}
string r;
for(int i=0;i<l;i++){
if(L[i]>R[i])
r += 'L';
else if(L[i]<R[i])
r += 'R';
else
r += '.';
}
cout<<r<<endl;
return 0;
}
发表于 2019-07-12 08:14:58
回复(0)
/*链接：https://www.nowcoder.com/questionTerminal/fae1307a24ae4e9ea852a646a4f812bf?answerType=1&f=discussion
来源：牛客网
这个字符串一共就分为四种情况：
(1)：L...L ，在这种情况下，将里面‘.’全部换成'L'；
(2)：R...L，在这种情况下，将里面左半部分替换为'L'，右半部分替换为'R'，需要注意的是区间长度的奇偶问题
(3)：R...R，在这种情况下，将里面‘.’全部换成'R'；
(4)：L...R，在这种情况下，我们不做任何处理；
备注：为了考虑左右开头的影响，我们分别在字符串的最左边加上“L”，最右边加上“R”，
这样既不会影响内部的判断，又能处理端点的情况
*/
import java.util.*;
public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String s = sc.nextLine();
StringBuilder sb = new StringBuilder(s);//为了统一处理，
sb.append("R");//在最右边加一个“R”
sb.insert(0, "L");//在最左边加一个“L”
List<Integer> v = new ArrayList<>();//记录含有L或R的下标位置
for(int i = 0; i < sb.length(); ++i){
if(sb.charAt(i) == 'L' || sb.charAt(i) == 'R')
v.add(i);;
}
for(int p = 0; p < v.size() - 1; ++p){//按RL区间端点分类，共有4类
int i = v.get(p), j = v.get(p + 1);//一个区间一个区间处理（区间都是相邻的处理）
char c1 = sb.charAt(i), c2 = sb.charAt(j);
if(c1 == 'L' && c2 == 'L'){
for(int k = i + 1; k < j; ++k)
sb.setCharAt(k, 'L');
}
if(c1 == 'L' && c2 == 'R'){}//这种什么也不用做
if(c1 == 'R' && c2 == 'L'){
for(int k = i + 1; k < (double)(j + i)/2; ++k)
sb.setCharAt(k, 'R');
for(int k = j - 1; k > (double)(j + i)/2; --k)
sb.setCharAt(k, 'L');
}
if(c1 == 'R' && c2 == 'R'){
for(int k = i + 1; k < j; ++k)
sb.setCharAt(k, 'R');
}
}
sb.deleteCharAt(sb.length() - 1);
sb.deleteCharAt(0);//把开始加的那两张去掉
System.out.println(sb);
}
}
发表于 2020-06-18 19:40:47
回复(0)
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text.RegularExpressions;
//总结目前牛客问题 第一,没有循环输入问题, 第二 有循环输入问题, 第三 输入有多余空格问题 ,第四 中间插入多余空行问题 ....
namespace Test0001
{
class Program
{
public static void Main(string[] args)
{
string line;
while (!string.IsNullOrEmpty(line = Console.ReadLine())) Func(line);
}
public static void Func(string line)
{
int n = line.Length;
int[] dp1 = new int[n];
int[] dp2 = new int[n];
//从左往右
dp1[0] = line[0] == 'R' ? n : 0; //初始值
for (int i = 1; i < n; i++)
{
if (line[i] == 'R')
{
dp1[i] = n;
}
else if (line[i] == '.')
{
dp1[i] = dp1[i - 1] > 0 ? dp1[i - 1] - 1 : 0;//如果左边的大于0说明是连续的,因此等于左边一位的值减1
}
}
//从右往左
dp2[n - 1] = line[n - 1] == 'L' ? n : 0;//初始值
for (int i = n - 2; i >= 0; i--)
{
if (line[i] == 'L')
{
dp2[i] = n;
}
else if (line[i] == '.')
{
dp2[i] = dp2[i + 1] > 0 ? dp2[i + 1] - 1 : 0;//如果右边的大于0说明是连续的,因此等于右边一位的值减1
}
}
for (int i = 0; i < n; i++)
{
int a = dp1[i] - dp2[i];
Console.Write(a > 0 ? 'R' : a < 0 ? 'L' : '.');
}
}
}
}
很巧妙的方法参考楼上数组差值做法
发表于 2019-12-06 17:44:25
回复(0)
#include <bits/stdc++.h>
using namespace std;
void get_L(vector<pair<int, int> > &p, int L, string s){
int power = 1;
p[L].first = power;
power++;
L--;
while(L >= 0 && s[L] != 'L' && s[L] != 'R'){
p[L].first = power;
power++;
L--;
}
}
void get_R(vector<pair<int, int> > &p, int R, string s){
int power = 1;
p[R].second = power;
power++;
R++;
while(R < s.length() && s[R] != 'L' && s[R] != 'R'){
p[R].second = power;
power++;
R++;
}
}
int main(int argc, char *argv[]){
string s;
cin >> s;
vector<pair<int, int> > p(s.length(), {0, 0});
for (int i = 0; i < s.length(); i++){
if(s[i] == 'L'){
get_L(p, i, s);
} else if(s[i] =='R'){
get_R(p, i, s);
}
}
for (int i = 0; i < s.length(); i++){
if(p[i].first > p[i].second){
if(p[i].second == 0){
s[i] = 'L';
} else {
s[i] = 'R';
}
} else if(p[i].first < p[i].second) {
if(p[i].second > p[i].first){
if(p[i].first == 0){
s[i] = 'R';
} else {
s[i] = 'L';
}
}
} else {
s[i] = '.';
}
}
for (int i = 0; i < s.length(); i++){
printf("%c", s[i]);
}
printf("\n");
return 0;
}
发表于 2019-09-16 16:33:18
回复(0)
stones = list(input())
prev = [-1, ""]
for i in range(len(stones)):
if stones[i] == "L":
if prev[1] == "L" or prev[1] == "":
for j in range(prev[0] + 1, i):
stones[j] = "L"
elif prev[1] == "R":
for j in range(prev[0] + 1, i):
if j < (prev[0] + i) / 2:
stones[j] = "R"
elif j > (prev[0] + i) / 2:
stones[j] = "L"
else:
continue
prev = [i, stones[i]]
elif stones[i] == "R":
if prev[1] == "R":
for j in range(prev[0] + 1, i):
stones[j] = "R"
prev = [i, stones[i]]
if prev[1] == "R":
for i in range(prev[0] + 1, len(stones)):
stones[i] = "R"
print("".join(stones))
发表于 2019-09-05 02:36:39
回复(0)
#include <iostream>
using namespace std;
int main(void){
enum {INIT, LEFT, RIGHT};
string s;
cin>>s;
int boundary = s.length();
int left = 0, right = 0, stage = INIT;
for(int i = 0; i < boundary; ++i){
if(s[i] == 'L'){
switch (stage)
{
//注意这里为了节省代码量，INIT case没有break
case INIT:
stage = LEFT;
case LEFT:
for(int j = left; j < i; ++j)
s[j] = 'L';
left = i+1;
break;
//(LEFT RIGHT)完成向中心倒的操作，此后状态回归到INIT
case RIGHT:
stage = INIT;
for(int j = right, k = i-1; j < k; ++j, --k)
s[j] = 'R', s[k] = 'L';
left = right = i+1;
}
}
if(s[i] == 'R'){
switch (stage)
{
case INIT:
case LEFT:
stage = RIGHT;
right = i+1;
break;
case RIGHT:
for(int j = right; j < i; ++j)
s[j] = 'R';
right = i+1;
}
}
}
//注意最后退出循环时若状态为RIGHT，完成向右倒的操作
if(stage == RIGHT)
for(int j = right; j < boundary; ++j)
s[j] = 'R';
cout<<s<<endl;
return 0;
} 使用状态机，分三种状态INIT，LEFT，RIGHT，初始状态为INIT，循环遍历字符串，遇到L或R字符进行相应的状态机转移和骨牌倾倒模拟操作，变量left，right表示向左向右倾倒时左边界值，倾倒时右边界值为i
若当前字符s[i]为L
| 当前状态 | 转移状态 | 操作 |
| INIT | LEFT | left至i的字符赋值为L，表示向左倾倒 |
| LEFT | LEFT(不变) | 同上 |
| RIGHT | INIT | right至i的字符从两端开始左右各赋值R，L，表示两边向中间倾倒 |
若当前字符s[j]为R
| 当前状态 | 转移状态 | 操作 |
| INIT | RIGHT | right=i+1 |
| LEFT | RIGHT(不变) | 同上 |
| RIGHT | RIGHT | right至i的字符赋值为R，表示向右倾倒 |
另外：对于RL中间插了一个R或L的情况，不才认为题意应该强调所有推倒是同时操作的
发表于 2019-08-16 11:35:46
回复(0)
从左到右，依次遍历
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String str = in.nextLine();
char[] c = str.toCharArray();
int i = -1;
int j = 0;
while (j < c.length){
if (c[j] == '.')
j++;
else if (c[j] == 'L'){
for (int k=i+1;k<j;k++){
c[k] = 'L';
}
i = j;
j++;
}else if (c[j] == 'R'){
int k;
for (k=j+1;k<c.length;k++){
if (c[k] != '.')
break;
}
j++;
if (k == c.length){
while (j < c.length){
c[j] = 'R';
j++;
}
}else if (c[k] == 'R'){
while (j < k){
c[j] = 'R';
j++;
}
i = k;
}else if (c[k] == 'L'){
int t = k-1;
while (j < t){
c[j] = 'R';
c[t] = 'L';
j++;t--;
}
j = k+1;
i = k;
}
}
}
for (char ch : c)
System.out.print(ch);
}
}
发表于 2019-08-06 16:42:46
回复(0)
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
char[] chs = scanner.next().toCharArray();
int len = chs.length;
int[] values = new int[len];
// 第一遍从左往右遍历，遇到R就赋值length，接下来如果是.则递减，直到遇到L清零
int value = 0;
for (int i = 0; i < len; i++) {
if (chs[i] == 'R') {
value = len;
} else if (chs[i] == 'L') {
value = 0;
} else {
// 保证L之后遇到.仍旧赋值0而不是负数
value = Math.max(value - 1, 0);
}
// values数组加上value值
values[i] += value;
}
// 第二遍从右往左遍历，遇到L就赋值length，接下来如果是.则递减，直到遇到R清零
value = 0;
for (int i = len - 1; i >= 0; --i) {
if (chs[i] == 'L') {
value = len;
} else if (chs[i] == 'R') {
value = 0;
} else {
// 保证R之后遇到.仍旧赋值0而不是负数
value = Math.max(value - 1, 0);
}
// values数组减去value值
values[i] -= value;
}
StringBuilder result = new StringBuilder();
// values值大于0则为R，小于0则为L，等于0则为.
for (int i : values) {
result.append(i > 0 ? 'R' : i < 0 ? 'L' : '.');
}
System.out.println(result);
}
}
编辑于 2019-07-02 19:39:26
回复(2)
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Solution10_推倒吧骨牌 {
public static void main(String[] args) throws IOException {
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
String s = bf.readLine();
int n = s.length();
int[] ltor = new int[n];//模拟往右边推，每个骨牌倒下的先后顺序
int[] rtol = new int[n];//模拟往左边推，每个骨牌倒下的先后顺序
boolean isPush = false;
for (int i = 0; i < n; i++) {
if (s.charAt(i) == 'R') {
ltor[i] = 1;
isPush = true;
} else if (s.charAt(i) == 'L') {
isPush = false;
} else if (isPush) {
ltor[i] = ltor[i - 1] + 1;
}
}
isPush = false;
for (int i = n - 1; i >= 0; i--) {
if (s.charAt(i) == 'L') {
isPush = true;
rtol[i] = 1;
} else if (s.charAt(i) == 'R') {
isPush = false;
} else if (isPush) {
rtol[i] = rtol[i + 1] + 1;
}
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < n; i++) {
if (ltor[i] == rtol[i]){
sb.append('.');
}else if (ltor[i] == 0 && rtol[i] != 0){//ltor[i] == 0，往左边倒
sb.append('L');
}else if (ltor[i] !=0 && rtol[i] == 0) {//rtol[i] = 0 往右边的倒
sb.append('R');
}else if (ltor[i] < rtol[i]){//往右边倒的时间小于往左边推到的时间，所以先往右边先倒
sb.append('R');
}else if (ltor[i]>rtol[i]){
sb.append('L');
}
}
System.out.println(sb.toString());
}
}
发表于 2019-08-15 22:06:28
回复(0)
import java.util.*;
public class TTTest {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String s = input.next();
input.close();
char[] a = s.toCharArray();
int len = s.length();
int lway = 0;
int ridx= 0;
int num = 0;
for (int i=0;i<len;i++) {
if (a[i] == 'R') {
for (int j=i+1;j<len;j++) {
if (a[j] == 'R') {
num = 1;
ridx = j;
break;
}
if (a[j] == 'L') {
num = 1;
lway = j;
if (((j-i)%2) != 0) {
for (int k=i+1;k<j;++k) {
if (k>((j-i)/2 + i)) {
a[k] = 'L';
}
if (k <= ((j-i)/2 + i)) {
a[k] = 'R';
}
}
}
if (((j-i)%2) == 0) {
for (int k=i+1;k<j;k++) {
if (k<((j+i)/2)) {
a[k] = 'R';
}
if (k>((j+i)/2)) {
a[k] = 'L';
}
if (k==((j+i)/2)) {
a[k] = '.';
}
}
}
break;
}
else {
num = 0;
}
}
if (num == 0) {
for (int j=i+1;j<len;j++) {
a[j] = 'R';
}
}
}
if(i< ridx){
a[i] = 'R';
}
if (a[i] == 'L') {
for (int j=lway;j<=i;j++) {
a[j] = 'L';
}
}
}
System.out.println(a);
}
} 很乱的解法，具体思路是 R 的话往右看，忽略掉 ‘ . ’ 最近是 R 的话中间的 ‘ . ’全部置为 R ，跳到左右边的 R ，如此类推；如果最近是 L ，则把中间置为 ‘RRLL’ 或 ‘RR . LL’ 这样的形式。这样遇到是 L 的时候，就把它左边到 上一个 ‘RR . LL’ 结尾的元素全部置为 L 。
发表于 2020-08-29 01:18:13
回复(0)
import sys from collections import deque instr = sys.stdin.readline().strip() q = deque() outstr = "" for ch in instr: if ch == '.': q.extend(ch) # `'.'`入队列 elif ch == 'R': if len(q) == 0: q.extend(ch) continue if q[0] == '.': outstr += ''.join(q) # 全部出队列 elif q[0] == 'R': outstr += 'R' * len(q) # 全部出队列，且被推倒 q.clear() q.extend(ch) # `'R'`入队列 elif ch == 'L': if len(q) == 0: outstr += 'L' continue if q[0] == '.': outstr += 'L' * len(q) # 全部出队列，且被推倒 q.clear() outstr += 'L' # 当前的`'L'`` elif q[0] == 'R': d, v = divmod(len(q) - 1, 2) if v: # 奇数个 outstr += 'R' + 'R' * d + '.' + 'L' * d + 'L' else: # 偶数个 outstr += 'R' + 'R' * d + 'L' * d + 'L' q.clear() if len(q) > 0: outstr += q[0] * len(q) print(outstr)
发表于 2020-06-09 23:08:21
回复(0)
感觉这样写比较好懂
#include "iostream"
#include "vector"#include "string"
using namespace std;
int main() {
string s;
cin >> s;
int stop = 0; // 若下一个非点值为L, 倾倒到此为止
for (int i = 0; i < s.size(); ++i) {
if (s[i] == 'L') {
for (int k = stop; k <= i; ++k) {
s[k] = 'L';
}
stop = i + 1;
}
else if (s[i] == 'R') {
int j = i + 1;
while (j < s.size() && s[j] == '.') {
j++;
}
if (j == s.size()||s[j]=='R') {
for (int k = i + 1; k < j; ++k) {
s[k] = 'R';
}
i = j-1;
}
else if (s[j] == 'L') {
int tmp = j;
while (i < j) {
s[i++] = 'R';
s[j--] = 'L';
}
if (i == j) {
s[i] = '.';
}
stop = tmp + 1;
i = tmp;
}
}
}
cout << s << endl;
system("pause");
return 0;
}
发表于 2020-06-02 17:06:03
回复(0)
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;
import java.util.Queue;
import java.util.Scanner;
import java.util.Stack;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
char[] starr = str.toCharArray();
//System.out.println(starr[1]);
int i, poi = 0;
char lesid = 'N';
int pl = -1;
StringBuilder sb = new StringBuilder();
while (poi < starr.length) {
if (starr[poi] != '.'||poi==starr.length-1) {
if (starr[poi] == 'L') {
if (lesid == 'N' || lesid == 'R') {
for (i = 0; i < (poi - pl - 1); i++) {
if((poi - pl - 1)%2==0) {
if(i<(poi - pl - 1)/2)
{
sb.append('R');
}else {
sb.append('L');
}
}else {
if(i<(poi - pl - 1)/2)
{
sb.append('R');
}else {
if(i==(poi - pl - 1)/2)
sb.append('.');
else
sb.append('L');
}
}
//sb.append('K');
}
} else if (lesid == 'L') {
for (i = 0; i < (poi - pl - 1); i++) {
sb.append('L');
}
}
sb.append('L');
pl = poi;
lesid = 'L';
} else if (starr[poi] == 'R') {
if(lesid == 'R') {
for (i = 0; i < (poi - pl - 1); i++) {
sb.append('R');
}
}else if(lesid == 'L') {
for (i = 0; i < (poi - pl - 1); i++) {
sb.append('.');
}
}
sb.append('R');
pl = poi;
lesid = 'R';
}
if(starr[poi] == '.'&&poi==starr.length-1) {
if(lesid == 'R') {
for (i = 0; i < (poi - pl ); i++) {
sb.append('R');}
}else {
for (i = 0; i < (poi - pl ); i++) {
sb.append('.');
}
}
}
}
poi++;
}
System.out.println(sb.toString());
}
}
import java.util.Arrays;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;
import java.util.Queue;
import java.util.Scanner;
import java.util.Stack;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
char[] starr = str.toCharArray();
//System.out.println(starr[1]);
int i, poi = 0;
char lesid = 'N';
int pl = -1;
StringBuilder sb = new StringBuilder();
while (poi < starr.length) {
if (starr[poi] != '.'||poi==starr.length-1) {
if (starr[poi] == 'L') {
if (lesid == 'N' || lesid == 'R') {
for (i = 0; i < (poi - pl - 1); i++) {
if((poi - pl - 1)%2==0) {
if(i<(poi - pl - 1)/2)
{
sb.append('R');
}else {
sb.append('L');
}
}else {
if(i<(poi - pl - 1)/2)
{
sb.append('R');
}else {
if(i==(poi - pl - 1)/2)
sb.append('.');
else
sb.append('L');
}
}
//sb.append('K');
}
} else if (lesid == 'L') {
for (i = 0; i < (poi - pl - 1); i++) {
sb.append('L');
}
}
sb.append('L');
pl = poi;
lesid = 'L';
} else if (starr[poi] == 'R') {
if(lesid == 'R') {
for (i = 0; i < (poi - pl - 1); i++) {
sb.append('R');
}
}else if(lesid == 'L') {
for (i = 0; i < (poi - pl - 1); i++) {
sb.append('.');
}
}
sb.append('R');
pl = poi;
lesid = 'R';
}
if(starr[poi] == '.'&&poi==starr.length-1) {
if(lesid == 'R') {
for (i = 0; i < (poi - pl ); i++) {
sb.append('R');}
}else {
for (i = 0; i < (poi - pl ); i++) {
sb.append('.');
}
}
}
}
poi++;
}
System.out.println(sb.toString());
}
}
发表于 2020-06-01 13:54:32
回复(0)
/*
LL LR RL RR
*/
import java.util.Scanner;
public class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
char[] in = sc.next().toCharArray();
int len = in.length;
char first = 'a';
int firsti = 0;
char second = 'a';
int secondi = 0;
int front = 0;
int pair = 0;
for (int i = 0; i < len; i++) {
if (in[i] == '.') {
continue;
}
else if (in[i] == 'L') {
if (pair == 0) {
first = 'L';
firsti = i;
pair++;
}
else {
second = 'L';
secondi = i;
pair++;
}
}
else {
if (pair == 0) {
first = 'R';
firsti = i;
pair++;
}
else {
second = 'R';
secondi = i;
pair++;
}
}
if (pair == 2) {
if (first == 'L' && second == 'L') {
pair = 1;
for (int j = front; j < secondi; j++) {
System.out.print('L');
}
front = secondi;
firsti = secondi;
}
if (first == 'L' && second == 'R') {
pair = 1;
for (int j = front; j <= firsti; j++) {
System.out.print('L');
}
for (int j = firsti+1; j < secondi; j++) {
System.out.print('.');
}
front = secondi;
first = 'R';
firsti = secondi;
}
if (first == 'R' && second == 'L') {
pair = 1;
for (int j = front; j < firsti; j++) {
System.out.print('.');
}
for (int j = firsti; j < firsti+(secondi-firsti+1)/2; j++) {
System.out.print('R');
}
int s = firsti+(secondi-firsti+1)/2;
if ((secondi-firsti+1)%2 == 1) {
System.out.print('.');
s++;
}
for (int j = s; j < secondi; j++) {
System.out.print('L');
}
front = secondi;
first = 'L';
firsti = secondi;
}
if (first == 'R' && second == 'R') {
pair = 1;
for (int j = front; j < firsti; j++) {
System.out.print('.');
}
for (int j = firsti; j < secondi; j++) {
System.out.print('R');
}
front = secondi;
firsti = secondi;
}
}
}
if (first == 'L') {
System.out.print('L');
front++;
for (int j = front; j < len; j++) {
System.out.print('.');
}
}
else {
for (int j = front; j < len; j++) {
System.out.print('R');
}
}
}
}
发表于 2020-05-31 09:54:38
回复(0)
#include<bits/stdc++.h>
using namespace std;
// 更新任意两个操作之间的骨牌
// 只有LR、LL、RR、RL这几种情况
void update(string& s,int l,int r,char last,char now)
{
if(last=='L'&&now=='R') return;
else if(last=='R'&&now=='L'){
while(r-l>2)
{
s[++l]='R';s[--r]='L';
}
}
else if(last=='L'&&now=='L'){
for(int i=l;i<=r;++i) s[i]='L';
}else{
for(int i=l;i<=r;++i) s[i]='R';
}
}
int main()
{
string s;
cin>>s;
bool first = true;
int i = 0;
int last_index = 0;
char pre;
while(i<s.size())
{
while(i<s.size()&&s[i]=='.') ++i;
// 先确定第一个动作是L还是R
if(s[i]=='L'){
if(first)
{
first = false;
for(int k=0;k<i;++k) s[k]='L';
pre = 'L';
last_index = i;
}
// 逐段更新
else {
update(s,last_index,i,pre,'L');
pre = 'L';
last_index = i;
}
}
else if(s[i]=='R'){
if(first){
first = false;
last_index = i;
pre = 'R';
}
else{
// 逐段更新
update(s,last_index,i,pre,'R');
pre = 'R';
last_index = i;
}
}
++i;
}
// 如果最后一个动作是R 后面几张牌改一下
if(pre=='R')
for(int k=last_index;k<s.size();++k) s[k]='R';
cout<<s<<endl;
return 0;
}
编辑于 2020-05-04 22:52:26
回复(0)
a = [0, 0] a = a + list(input()) a = a + [0, 0] s = '' for i in range(2, len(a)-2): if a[i] == '.' and a[i-1] == 'R': c = 1 for j in range(i+1, len(a)-2): if a[j] == '.': c = c + 1 continue if a[j] == 'R': break if a[j] == 'L': for k in range(1, c+1): if k <= c//2: a[i+k-1] = -1 if c % 2 == 0 and k > c//2: a[i+k-1] = 1 if c % 2 != 0 and k > c//2 + 1: a[i+k-1] = 1 if c % 2 != 0 and k == c//2 + 1: a[i+k-1] = 0 break for i in range(2, len(a)-2): if a[i] == 'R' and a[i+1] == '.' and a[i+2] != 'L': a[i+1] = 'R' for i in range(1, len(a)-3): if a[-i] == 'L' and a[-i-1] == '.' and a[-i-2] != 'R': a[-i-1] = 'L' for i in range(2, len(a)-2): if a[i] == 0: a[i] = '.' if a[i] == -1: a[i] = 'R' if a[i] == 1: a[i] = 'L' s = s + a[i] print(s)
1、确定R及L中间点的个数，把点变为-1（R），0（.），1（L）
2、从左到右遍历，把R右边的点变为R
3、从右到左遍历，把L左边的点变为L
编辑于 2020-05-03 09:29:14
回复(0)
问题信息
通过挑战的用户
查看代码-
2022-10-20 20:29:28 -
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2022-09-29 16:05:40
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