Find the Duplicate Number

题面：

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:

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

题解：原本想快排下，两个循环若有一样的值，便是所求。

class Solution {public:    int findDuplicate(vector<int>& nums) {        int i;        int j;        sort(nums.begin(),nums.end());        for(i=0;i<nums.size();i++){           for(j=i+1;j<nums.size();j++){               if(nums[i]==nums[j])               return nums[i];           }        }    }};

惊讶的是也能过？后来看到条件1不能改变数组。就想到2分查找。

int findDuplicate(vector<int>& nums) {    int n=nums.size()-1;    int low=1;    int high=n;    int mid;    while(low<high){        mid=(low+high)/2;        int count=0;        for(int num:nums){            if(num<=mid) count++;        }        if(count>mid) high=mid;        else low=mid+1;     }    return low;}