[编程题]打印回形数
import java.util.Scanner;
/**
* 欢聚时代2018校招笔试题-Android B卷 打印回形数
* 回形数是一个矩阵输出，从矩阵的左上角往右开始打印数字0，遇到矩阵边界时，顺时针90方向
* 继续打印，并数字增长1，如此类推直到把矩阵填满，输入一个整形宽和高单位，每输出一个数
* 字，占用1单位宽高空间，根据入参打印出对应的回形数
* 输入描述:
* 矩阵的宽高
* 输出描述:
* 回字形的矩阵
* 输入例子1:
* 8 3
* 输出例子1:
* 00000000
* 34444441
* 22222221
*/
public class PrintBackMatrix {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int col = sc.nextInt();
int row = sc.nextInt();
printMatrix(col, row);
}
public static void printMatrix(int columns, int rows) {
int[][] matrix = new int[rows][columns];
int start = 0;
while (columns > start * 2 && rows > start * 2) {
printMatrixInCircle(matrix, columns, rows, start);
start++;
}
int m = matrix.length;
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
System.out.print(matrix[i][j]);
}
System.out.println();
}
}
public static void printMatrixInCircle(int[][] matrix, int columns, int rows,
int start) {
int endX = columns - 1 - start;
int endY = rows - 1 - start;
int num = 0;
//第一步：从左到右打印一行
for (int i = start; i <= endX; i++) {
matrix[start][i] = start * 4 + num;
}
num++;
//第二步：从上到下打印一列
if (start < endY) {
for (int i = start + 1; i <= endY; i++) {
matrix[i][endX] = start * 4 + num;
}
}
num++;
//第三步：从右到左打印一行
if (start < endX && start < endY) {
for (int i = endX - 1; i >= start; i--) {
matrix[endY][i] = start * 4 + num;
}
}
num++;
//第四步：从下到上打印一列
if (start < endX && start < endY - 1) {
for (int i = endY - 1; i >= start + 1; i--) {
matrix[i][start] = start * 4 + num;
}
}
num = 0;
}
}
发表于 2018-07-09 21:20:36
回复(0)
入门题。。
import java.util.*;
public class Main {
private static int MAX = (int) 10e5+5;
private static int INT_MAX = 0x3f3f3f3f;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int m = sc.nextInt(), n = sc.nextInt();
int count = 0;
int[][] map = new int[n][m];
for (int i=0; i!=n; i++) {
for (int j=0; j!=m; j++) {
map[i][j] = -1;
}
}
int i = 0, j = -1; int now = 0;
while (count < m * n) {
while (j+1<m && map[i][j+1]==-1 && count < m*n) { map[i][++j] = now; count++; } now++;
while (i+1<n && map[i+1][j]==-1 && count < m*n) { map[++i][j] = now; count++; } now++;
while (j-1>=0 && map[i][j-1]==-1 && count < m*n) { map[i][--j] = now; count++; } now++;
while (i-1>=0 && map[i-1][j]==-1 && count < m*n) { map[--i][j] = now; count++; } now++;
}
StringBuilder sb = new StringBuilder();
for (i=0; i!=n; i++) {
for (j=0; j!=m; j++) {
sb.append(map[i][j]);
}
sb.append("\n");
}
System.out.print(sb.toString());
return;
}
}
编辑于 2019-02-11 01:34:05
回复(0)
直接上模拟边界法
import java.lang.String;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line;
while((line = br.readLine()) != null) {
String[] strWH = line.split(" ");
int h = Integer.parseInt(strWH[1]), w = Integer.parseInt(strWH[0]);
solve(h, w);
}
}
private static void solve(int m, int n) {
int[][] arr = new int[m][n];
int idx = 0, start = 0;
int left = 0, right = n - 1, top = 0, bottom = m - 1;
while(idx <= m * n){
for(int i = left; i <= right; i++){
arr[top][i] = start;
idx ++;
}
top ++;
if(top > bottom) break;
start ++;
for(int i = top; i <= bottom; i++){
arr[i][right] = start;
idx ++;
}
right --;
if(left > right) break;
start ++;
for(int i = right; i >= left; i--){
arr[bottom][i] = start;
idx ++;
}
bottom --;
if(top > bottom) break;
start ++;
for(int i = bottom; i >= top; i--){
arr[i][left] = start;
idx ++;
}
left ++;
if(left > right) break;
start ++;
}
StringBuilder sb;
for(int i = 0; i < m; i++){
sb = new StringBuilder();
for(int j = 0; j < n; j++)
sb.append(arr[i][j]);
System.out.println(sb.toString());
}
}
}
编辑于 2020-11-30 17:08:01
回复(0)
public class Main { //一个时间复杂度O(width*height)，空间复杂度O(1)的算法
public static void main(String[] args) {
java.util.Scanner sc = new java.util.Scanner(System.in);
int width = sc.nextInt(), height = sc.nextInt(), i, j, minDis, prt;
for (i = 1; i <= height; ++i) {
for (j = 1; j <= width; ++j) {//本循环内只有判断结构和顺序结构，
minDis = i - 1; //所以循环内花费的时间是常数，
prt = 4 * minDis; //总的时间花费是常数*width*height，
if (width - j < minDis) { //略去常数，复杂度为O(width*height)。
minDis = width - j; //显然因为要打印width*height个字符，
prt = 1 + 4 * minDis; //所以不考虑常数的时间复杂度已最佳。
} //而因为没有使用一个矩阵存储，
if (height - i < minDis) { //空间复杂度只为常数，即O(1)
minDis = height - i;
prt = 2 + 4 * minDis;
}
if (j - 1 < minDis)
prt = 3 + 4 * (j - 1);
System.out.print(prt);
}
System.out.println("");
}
}
}
编辑于 2018-07-10 00:21:45
回复(1)
[cols,rows]=list(map(int,input().strip().split()))
res=[[0 for i in range(cols)]for i in range(rows)]
left=0
right=cols-1
up=0
down=rows-1
val=0
while left<=right and up<=down:
for j in range(left,right+1):
res[up][j]=val
val+=1
for i in range(up+1,down+1):
res[i][right]=val
if up<down:
val+=1
for i in range(right-1,left-1,-1):
res[down][i]=val
if left<right:#注意
val+=1
for j in range(down-1,up,-1):
res[j][left]=val
val+=1
left+=1
right-=1
up+=1
down-=1
for i in res:
print(''.join(list(map(str,i)))) 注意倒着走的两个单排防止回流的判断
发表于 2020-03-27 14:00:16
回复(0)
#include<iostream>
#include<string>
#include<vector>
using namespace std;
int main()
{
int N,M;
cin>>N>>M;
vector<vector<int>> rec(M);
vector<int> temp(N,-1);
for(int i=0;i<M;i++)
{
rec[i]=temp;
}
int x=0,y=0;
int value=0;
int pM=M,pN=N;
while(pM!=0&&pN!=0)
{
for(int i=0;i<pN;i++)
{
rec[x][y++]=value;
}
y--;
pM--;value++;
if(pM==0) break;
for(int i=0;i<pM;i++)
{
rec[++x][y]=value;
}
pN--;value++;
if(pN==0) break;
for(int i=0;i<pN;i++)
{
rec[x][--y]=value;
}
pM--;value++;
if(pM==0) break;
for(int i=0;i<pM;i++)
{
rec[--x][y]=value;
}
pN--;value++;
y++;
}
for(auto r:rec)
{
for(auto s:r)
cout<<s;
cout<<endl;
}
}
发表于 2019-09-10 15:12:12
回复(0)
#include<bits/stdc++.h>
// 剑指offer顺时针打印矩阵类似题目
using namespace std;
void printMatrixInCircle(vector<vector<int>> &res, int start, int width, int length){
int endy = length - start -1; //终止列号
int endx = width - start - 1; //终止行号
for(int i = start; i <= endy; i++)
res[start][i] = start*4;
if(endx > start){ //终止行号大于起始行号
for(int i = start+1; i <= endx; i++){
res[i][endy] = start*4 +1;
}
}
if(endx > start && endy > start){//终止行号大于起始行号 and 终止列号也大于起始列号
for(int i = endy-1; i >= start ; i--){
res[endx][i] = start*4 +2;
}
}
if(endx > start+1 && endy > start){//终止行号比起始行号大2或者以上 and 终止列号也大于起始列号
for(int i = endx-1; i > start ; i--){
res[i][start] = start*4 +3;
}
}
}
int main(){
int length,width;
cin>>length>>width;
if(length <= 0 || width <= 0)
return 0;
int start = 0;
vector<vector<int>> res(width, vector<int>(length, -1));
while(length > start*2 && width > start*2){
printMatrixInCircle(res, start++, width, length);
}
for(int i = 0; i < width; i++){
for(int j = 0; j< length; j++)
cout<<res[i][j];
cout<<endl;
}
return 0;
}
发表于 2019-09-05 09:24:03
回复(0)
只做越界判断，越界就走下一个方向
#include<bits/stdc++.h>
using namespace std;
int main(){
int m=0, n=0;
scanf("%d%d", &n, &m);
vector<vector<int>> mat(m, vector<int>(n, -1));
int direct[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};
int cnt=0, flag=0, x=0, y=0, cur=0;
while(cnt < n*m){
//如果没越界，一直走
while(0<=x&&x<m && 0<=y&&y<n && mat[x][y]==-1){
++cnt;
mat[x][y] = cur;
x += direct[flag][0];
y += direct[flag][1];
}
//越界了，往回走一步
x -= direct[flag][0];
y -= direct[flag][1];
//走下一个方向
cur += 1;
flag = (flag+1)%4;
x += direct[flag][0];
y += direct[flag][1];
}
for(int i=0; i<m; ++i){
for(const auto&x: mat[i]){
cout<< x;
}cout<< endl;
}
return 0;
}
发表于 2019-08-13 21:54:36
回复(0)
#include <bits/stdc++.h>
using namespace std;
void A (int rows,int columns,vector<vector<int>> matrix,int value, int start){
int endX=columns-1-start;
int endY=rows-1-start;
for(int i=start;i<=endX;i++){
matrix[start][i]=value;
value++;
}
if(start<endY){
for(int i=start+1;i<=endY;i++){
matrix[i][endX]=value;
value++;
}
}
if(start<endX&&start<endY){
for(int i=endX;i>=start;i--){
matrix[endY][i]=value;
value++;
}
}
if(start<endX&&start<endY-1){
for(int i=endY-1;i>=start+1;i--){
matrix[i][start]=value;
value++;
}
}
}
int main (){
int n,m;
vector <vector<int>> M;
while(cin>>n>>m){
int val=0;
int sta=0;
while (n > sta * 2 && m > sta * 2) {
A ( n,m,M[n][m], val, sta);
sta++;
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cout<<M[i][j];
}
cout<<endl;
}
}
为啥数组越界~
编辑于 2019-07-10 19:38:48
回复(0)
# -*- coding: utf-8 -*- """ Created on Sun May 12 10:34:14 2019 @author: i """ def mawari(a,h1,h2,w1,w2,start): if h2-h1 == 0: for j in range(w1,w2+1): a[h1][j] = start return if w2-w1 == 0: a[h1][w1] = start for i in range(h1+1,h2+1): a[i][w1] = start+1 return if h2-h1 == 1: for j in range(w1,w2+1): a[h1][j] = start a[h2][w2] = start+1 for j in range(w1,w2): a[h2][j] = start+2 return if w2-w1 == 1: for j in range(w1,w2+1): a[h1][j] = start for i in range(h1+1,h2+1): a[i][w2] = start+1 a[h2][w1] = start+2 for i in range(h1+1,h2): a[i][w1] = start+3 return for j in range(w1,w2+1): a[h1][j] = start for i in range(h1+1,h2+1): a[i][w2] = start+1 for j in range(w1,w2): a[h2][j] = start+2 for i in range(h1+1,h2): a[i][w1] = start+3 mawari(a,h1+1,h2-1,w1+1,w2-1,start+4) if __name__ == "__main__": line = raw_input() lines = line.split() W,H = int(lines[0]),int(lines[1]) a = [[0 for j in range(W)] for i in range(H)] mawari(a,0,H-1,0,W-1,0) for i in range(H): result = [str(item) for item in a[i]] print "".join(result)
发表于 2019-05-12 20:01:34
回复(0)
import java.util.Scanner;
/**
* 详见《剑指offer》 顺时针打印矩阵
* https://blog.csdn.net/qq_32767041/article/details/83831330#421__1344
*
* @author wylu
*/
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()) {
int cols = scanner.nextInt(), rows = scanner.nextInt();
int[][] matrix = new int[rows][cols];
int value = 0, start = 0;
while (cols > start * 2 && rows > start * 2) {
value = calMatrixInCircle(matrix, rows, cols, start++, value);
}
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
System.out.print(matrix[i][j]);
}
System.out.println();
}
}
}
private static int calMatrixInCircle(int[][] matrix,
int rows, int cols,
int start, int value) {
int endX = cols - start - 1, endY = rows - start - 1;
for (int i = start; i <= endX; i++) matrix[start][i] = value;
value++;
if (endY > start) {
for (int i = start + 1; i <= endY; i++) matrix[i][endX] = value;
value++;
}
if (endY > start && endX > start) {
for (int i = endX - 1; i >= start; i--) matrix[endY][i] = value;
value++;
}
if (endX > start && endY - start > 1) {
for (int i = endY - 1; i > start; i--) matrix[i][start] = value;
value++;
}
return value;
}
}
发表于 2019-01-10 17:30:18
回复(0)
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