列表

详情


NC283. 数组中重复的数字

描述

在一个长度为n的数组里的所有数字都在0到n-1的范围内。 数组中某些数字是重复的,但不知道有几个数字是重复的。也不知道每个数字重复几次。请找出数组中任意一个重复的数字。 例如,如果输入长度为7的数组[2,3,1,0,2,5,3],那么对应的输出是2或者3。存在不合法的输入的话输出-1

数据范围:
进阶:时间复杂度,空间复杂度

示例1

输入:

[2,3,1,0,2,5,3]

输出:

2

说明:

2或3都是对的

原站题解

import java.util.*;
public class Solution {
/**
*
*
*
* @param numbers int
* @return int
*/
public int duplicate (int[] numbers) {
// write code here
}
}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

C++ 解法, 执行用时: 3ms, 内存消耗: 640KB, 提交时间: 2021-10-30

class Solution {
public:
/**
*
*
*
* @param numbers intvector
* @return int
*/
int duplicate(vector<int>& numbers) {
// write code here
int len = numbers.size();
for(int i = 0 ; i < len ;++i)
{
while( numbers[i] != i )
{
int t = numbers[i];
if( numbers[t] != t )
{
numbers[i] = numbers[t];
numbers[t] = t;
}
else return t;
}
swap(numbers[i],numbers[numbers[i]]);
}
return -1;
}
};

C++ 解法, 执行用时: 3ms, 内存消耗: 644KB, 提交时间: 2021-09-07

class Solution {
public:
/**
*
*
*
* @param numbers intvector
* @return int
*/
int duplicate(vector<int>& numbers) {
// write code here
int i = 0;
while(i < numbers.size()) {
if(numbers[i] == i) {
i++;
continue;
}
if(numbers[numbers[i]] == numbers[i])
return numbers[i];
swap(numbers[i],numbers[numbers[i]]);
}
return -1;
}
};

C++ 解法, 执行用时: 3ms, 内存消耗: 648KB, 提交时间: 2021-11-01

class Solution {
public:
/**
*
*
*
* @param numbers intvector
* @return int
*/
int duplicate(vector<int>& numbers) {
if(numbers.size() <= 1) return -1;
vector <int> v(numbers.size(),0);
for(int i = 0; i < numbers.size(); i++){
if(v[numbers[i]] != 0) return numbers[i];
v[numbers[i]] = -1;
}
return -1;
}
};

C++ 解法, 执行用时: 3ms, 内存消耗: 648KB, 提交时间: 2021-09-04

class Solution {
public:
/**
*
*
*
* @param numbers intvector
* @return int
*/
int duplicate(vector<int>& numbers) {
// write code here
sort(numbers.begin(),numbers.end());
int n = numbers.size();
int ans;
if(n<=0) return -1;
for(int i=0;i<n;i++)
{
if(numbers[i]==numbers[i+1])
ans = numbers[i];
}
return ans;
}
};

Rust 解法, 执行用时: 3ms, 内存消耗: 652KB, 提交时间: 2021-11-29

struct Solution{
}
impl Solution {
fn new() -> Self {
Solution{}
}
/**
*
*
*
* @param numbers int
* @return int
*/
pub fn duplicate(&self, numbers: Vec<i32>) -> i32 {
// write code here
if numbers.len() == 0 {
return -1;
}
let mut nums:Vec<i32> = numbers;
for i in 0..nums.len() {
while nums[i] != i as i32 {
if nums[i] == nums[nums[i] as usize] {
return nums[i];
}
let tmp = nums[i];
nums[i] = nums[nums[i] as usize];
nums[tmp as usize] = tmp;
}
}
return -1;
}
}

上一题