题目描述
找出数组中重复的数字。
在一个长度为 n 的数组 nums 里的所有数字都在 0~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重复了,也不知道每个数字重复了几次。请找出数组中任意一个重复的数字。
示例 1:
输入:
[2, 3, 1, 0, 2, 5, 3]
输出:2 或 3
限制:
2 <= n <= 100000
Java
import java.util.HashSet;
import java.util.Set;
class Solution {
public static void main(final String[] args) {
final Solution solution = new Solution();
final int[] nums = new int[] { 2, 3, 1, 0, 2, 5, 3 };
System.out.println(solution.findRepeatNumber(nums));
System.out.println(solution.findRepeatNumber2(nums));
}
/**
* 方法一:遍历数组
*
* 遍历数组中的每个元素: * 将该元素加入集合中,判断是否加入成功 * 添加失败,则找到重复数字
*
* 时间复杂度 O(n) 空间复杂度 O(n)
*
* @param nums
* @return
*/
public int findRepeatNumber(final int[] nums) {
Set<Integer> set = new HashSet<>();
int repeat = -1;
for (final int num : nums) {
if (!set.add(num)) {
repeat = num;
break;
}
}
return repeat;
}
/**
* 方法二:原地置换
* 注意所有数字都在 0~n-1 的范围内
* @param nums
* @return
*/
public int findRepeatNumber2(final int[] nums) {
int len = nums.length;
int temp = 0;
for (int i = 0; i < len; i++) {
if (nums[i] != i) {
if (nums[i] == nums[nums[i]]) {
return nums[i];
} else {
temp = nums[i];
nums[i] = nums[temp];
nums[temp] = temp;
}
}
}
return -1;
}
}
C++
#include<iostream>
#include<vector>
using namespace std;
class Solution {
public:
int findRepeatNumber(vector<int>& nums) {
int len = nums.size();
int temp = 0;
for (int i = 0; i < len; ++i) {
if (nums[i] != i) {
if (nums[i] == nums[nums[i]]) {
return nums[i];
} else {
temp = nums[i];
nums[i] = nums[temp];
nums[temp] = temp;
}
}
}
return -1;
}
};
int main() {
auto s = new Solution();
vector<int> v = {2, 3, 1, 0, 2, 5, 3};
cout << s->findRepeatNumber(v) << endl;
delete s;
}
Python
from __future__ import annotations
from typing import List
class Solution:
def findRepeatNumber(self, nums: List[int]) -> int:
size = len(nums)
temp = 0
for index, value in enumerate(nums):
if value != index:
if value == nums[value]:
return value
else:
temp = value
value = nums[temp]
nums[temp] = temp
return -1
if __name__ == "__main__":
s = Solution()
print(s.findRepeatNumber([2, 3, 1, 0, 2, 5, 3]))