287. Find the Duplicate Number

问题：

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.

答案：

n+1个1到n之间的整数，只有一个重复的数的话，要求找出这个唯一重复的数，满足如下条件：

1.不能改变这个数组（比如不能重新排序）

2.只能用连续的空间复杂度为o(1)的空间

3.时间复杂度小于o(n²)

4.虽然只有一个重复的数，但这个数可以重复多次

  思路：

    利用折半查找的思想，因为是1到n之间的数，令low=1，high=n,先求得一个中间值mid=(low+high)/2，遍历一遍数组，求出小于mid的元素的个数c，若c小于等于mid，则重复的数大于mid，令left=mid+1，重复以上步骤，反之则小于等于mid，令right=mid，重复以上步骤。随着范围的缩小，当low>=high时，可求出重复的数。

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
      int n = nums.size()-1;  
        int low = 1, high= n;  
        int mid = 0;  
        while(low<high) {  
            mid = (low+high)/2;  
            int c= count(nums, mid); 
            if(c<=mid) {  
                low = mid+1;  
            } else {  
                high = mid;  
            }  
        }  
        return low;  
    }  
      
    int count(vector<int>& nums, int mid) {  
        int c =0;  
        for(int i=0; i<nums.size(); i++) {  
            if(nums[i]<=mid) c++;  
        }  
        return c;    
    }
};