LeetCode 287. Find the Duplicate Number

题目：
Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

Note:

You must not modify the array (assume the array is read only).
You must use only constant, O(1) extra space.
Your runtime complexity should be less than O(n2).
There is only one duplicate number in the array, but it could be repeated more than once.

思路：
具体思路见注释。

代码1：

class Solution {
public:int findDuplicate(vector<int>& nums) {int n=nums.size()-1;int low=1;int high=n;int mid;while(low<high){//当low比high小时mid=(low+high)/2;//计算midint count=0;for(int i=0;i<=n;++i){//计算nums中比mid小于等于的数的数量 if(nums[i]<=mid){count++;}}if(count>mid){//如果count比mid大，high赋为midhigh=mid;}else{//否则low赋为mid+1low=mid+1;}  }return low;//循环结束，返回的low即为重复数字}
};

输出结果： 12ms

代码2：

class Solution {
public:int findDuplicate(vector<int>& nums) {int len = nums.size();for (int i = 0; i<len - 1; ++i){for (int j = i + 1; j<len; ++j){if (nums[i] == nums[j]){//暴力方法直接两层循环return nums[i];}} }return nums[low];}
};

输出结果： 422ms

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