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

暴力解：O（n^2)
二分搜索：统计大于mid的个数，如果比mid多则重复的在mid右边，反之相反：

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;
}

用环来做，追赶问题