[leetcode] 287. Find the Duplicate Number 解题报告

题目链接：https://leetcode.com/problems/find-the-duplicate-number/

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：这题有一个定理叫做鸽笼原理，大意就是n个东西放到m个容器中，如果n > m，那么必然有一个容器包含多于一个东西。

而这题把数据范围限制在1-n之间也是别有用意，这样我们就可以用二分查找来不断缩小范围来找到重复的数。

我们使用二分查找先确定一个中间值mid，然后统计整个数组，看比mid小的数是否比mid多，如果多的话，说明重复的值就在[left, mid-1]之间，否则就在[mid+1, right]之间。

代码如下：

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int len = nums.size();
        int left = 1, right = len;
        while(left <= right)
        {
            int mid = (left+right)/2, cnt = 0;
            for(int i = 0; i< len ; i++)
                if(nums[i] <= mid) 
                    cnt++;
            if(cnt <= mid)
                left = mid +1;
            else
                right = mid-1;
        }
        return left;
    }
};

参考：http://www.cnblogs.com/grandyang/p/4843654.html