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:

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.

思路：看到这个题目定义想法就是一个个找，但这样时间复杂就是O(n2)， 与题目要求不符。然后就想到可以用类似二分查找的思路来求解，先另low = 1,high = n,看重复的数字是出现在1到n中较小的一半还是出现在较大的一半，然后再从相应的一半按装这样找法，直到low==high时，就找到这个数字。

代码：

class Solution {
public:
    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 = 0;num<nums.size();num++){
                if(nums[num]<=mid) count++;
            }
            if(count>mid) high=mid;
            else low=mid+1; 
        }
        return low;
        }
};