给定一个长度为n的数组a[0],a[1]...a[n-1]和一个数字x，在数组中查找两个数a和b（可以是相同的值，但是不可以是相同位置的数字），使得它们的和与输入的数字差的绝对值最小。
比如：
a = [8,3,6,1] x=13
那么答案为：6和8
[编程题]数字查找
public static int[] Findmincha(int[] a ,int x) {
int[] res = new int[2];
Arrays.sort(a);
//System.out.println("x:"+x);
// for (int i = 0; i <a.length; i++) {
// System.out.println("a["+i+"]:"+a[i]);
// }
int min = Integer.MAX_VALUE;
for (int i = 0; i < a.length; i++) {
for (int j = i+1; j < a.length; j++) {
if (Math.abs(a[i]+a[j]-x)<min) {
min = Math.abs(a[i]+a[j]-x);
res[0]=a[i];
res[1]=a[j];
}
}
}
//System.out.print(res[0]+" "+res[1]);
return res;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int x = sc.nextInt();
int[] a =new int[n];
for (int i = 0; i < a.length; i++) {
a[i]=sc.nextInt();
}
int[] res = Findmincha(a,x);
System.out.print(res[0]+" "+res[1]);
} 这题只能一种答案，如果有两种答案的话，还不能选另一种？比如 3+4=7，2+5=7，但是选了另一种会报错，所以我的代码没有通过
发表于 2020-02-11 20:42:46
回复(0)
双指针对有序数组查找指定值，小改即可（但是测试用例有问题答案不唯一，不能通过）
n, x = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
st, ed = 0, n-1
a.sort()
temp = float('inf')
ans = []
while st<ed:
now = a[st]+a[ed]
if now>x:
if now-x<temp:
temp=now-x
ans = [a[st],a[ed]]
ed-=1
elif now<x:
if x-now<temp:
temp=x-now
ans = [a[st],a[ed]]
st+=1
else:
ans = [a[st],a[ed]]
break
print(*ans)
发表于 2023-08-08 00:31:55
回复(0)
import sys
a = [[0],[0]]
i = 0
for line in sys.stdin:
# print(line)
# a = line.strip('\n')
b = line.split('\n')[0]
b = b.split(' ')
a[i] = b
i += 1
tmp = {}
for i in range(int(a[0][0])):
for j in range(i+1, int(a[0][0])):
tmp[int(a[1][i])+int(a[1][j])] = (a[1][i], a[1][j])
tar1 = int(a[0][1])
tar2 = tar1
x = '0'
y = '0'
while 1:
if tar1 in tmp:
x, y = tmp[tar1]
break
if tar2 in tmp:
x, y = tmp[tar2]
break
else:
tar1 = tar1+1
tar2 = tar2-1
l = []
l.append(int(x))
l.append(int(y))
l.sort()
# print(l)
print(l[0],l[1]) 坑p问题，明明没说有多种情况怎么办，还答案只有一种情况，有多种情况时输出其他情况就算错，又不说清楚选择的标准是啥，我怎么知道什么算对什么算错
发表于 2023-03-27 22:37:15
回复(1)
#include <iostream>
#include <vector>
#include <climits>
#include <algorithm>
using namespace std;
int main() {
int n, target;
cin >> n >> target;
vector<int> nums(n);
for (int i = 0; i < n; i++) cin >> nums[i];
sort(nums.begin(), nums.end());
int greater = INT_MAX, smaller = INT_MIN;
int gArr[2], sArr[2];
// 双指针
int left = 0, right = n - 1;
while (left < right) {
int sum = nums[left] + nums[right];
if (sum > target) {
if (sum < greater) {
greater = sum;
gArr[0] = nums[left];
gArr[1] = nums[right];
}
right--;
}
else if (sum < target) {
if (sum > smaller) {
smaller = sum;
sArr[0] = nums[left];
sArr[1] = nums[right];
}
left++;
}
else {
printf("%d %d", nums[left], nums[right]);
return 0;
}
}
if (abs(smaller - target) < abs(greater - target)) printf("%d %d", sArr[0], sArr[1]);
else printf("%d %d", gArr[0], gArr[1]);
return 0;
}
// 64 位输出请用 printf("%lld")
// 答案不止一组，输出另一组还报错，就离谱
编辑于 2023-01-25 01:44:24
回复(0)
# str1=input()
# str2=input()
input1=str(input()).split(' ')
n,x=int(input1[0]),int(input1[1])
number1 = str(input()).split(' ')
number=[int(item) for item in number1]
num_abs = {}
for i in range(n-1):
for j in range(i+1, n):
list1=[]
append_num = abs(number[i]+number[j]-x)
if number[i]>number[j]:
list1.append(number[j])
list1.append(number[i])
else:
list1.append(number[i])
list1.append(number[j])
num_abs[append_num]=list1
# print(num_abs)
min_num=min(num_abs.keys())
# print(min_num)
# print(type(num_abs[min_num]))
min_value1=num_abs[min_num]
min_value=[str(item) for item in min_value1]
#print(min_value=num_abs[min_num])
print(' '.join(min_value))
编辑于 2022-06-21 16:30:42
回复(0)
n,x = list(map(int, input().split(' ')))
A = [int(x) for x in input().split(' ')]
num = len(A)
a = 0
b = 1
for i in range(num - 1):
for j in range(i+1, num):
if abs(A[i] + A[j] - x) < abs(A[a] + A[b] - x):
a = i
b = j
if A[a] < A[b]:
print(A[a], A[b])
else:
print(A[b], A[a])
A = [int(x) for x in input().split(' ')]
num = len(A)
a = 0
b = 1
for i in range(num - 1):
for j in range(i+1, num):
if abs(A[i] + A[j] - x) < abs(A[a] + A[b] - x):
a = i
b = j
if A[a] < A[b]:
print(A[a], A[b])
else:
print(A[b], A[a])
发表于 2020-06-10 22:54:15
回复(0)
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.StringTokenizer;
public class Main {
static int[] array1;
public static void main(String[] args) {
//输入两个数 数组大小n 要比较的数x
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int x = sc.nextInt();
//数组array1
if (2 <= n && n <= 1000 && 1 <= x && x <= 10E8) {
array1 = new int[n];
for (int i = 0; i < n; i++) {
int b = sc.nextInt();
array1[i] = b;
}
//结果放进map
Map map = new HashMap();
int c;
int e=1000000;
for (int i = 0; i < n; i++) {
for (int j = 1; j < n; j++) {
//去重
if (i == j || i > j) {
continue;
}
c = Math.abs(array1[i] + array1[j]- x);
map.put(c, Math.min(array1[i],array1[j]) + "," +Math.max(array1[i],array1[j]) );
e = Math.min(e, c);
}
}
// System.err.println("最小绝对值"+e);
StringTokenizer splitor = new StringTokenizer((String) map.get(e),",");
while(splitor.hasMoreTokens()){
System.out.print(splitor.nextElement()+" ");
}
}
}
}
发表于 2020-02-19 19:40:38
回复(0)
问题信息
通过挑战的用户
-
2022-11-17 15:41:32
-
2022-10-10 18:10:02 -
2022-10-10 17:50:58
-
2022-08-30 20:50:28
-
2022-08-08 12:03:11
相关试题
-
扫描二维码，关注牛客网
- 意见反馈
-
下载牛客APP，随时随地刷题
扫一扫，把题目装进口袋
- 求职之前，先上牛客
-
扫描二维码，进入QQ群
扫描二维码，关注牛客公众号
- 公司地址：北京市朝阳区北苑路北美国际商务中心K1座一层-北京牛客科技有限公司
- 联系方式：010-60728802 投诉举报电话：010-57596212（朝阳人力社保局）
- 牛客科技© All rights reserved admin@nowcoder.com
- 京ICP备14055008号-4 增值电信业务经营许可证 营业执照 人力资源服务许可证
-
京公网安备
11010502036488号
