[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:

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：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 left = 1, right = nums.size()-1;
while(left <= right)
{
int cnt = 0, mid = (left+right)/2;
for(auto val: nums)
if(val <= mid) cnt++;
if(cnt <= mid) left = mid+1;
else right = mid-1;
}
return left;
}
};

<span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">参考：http://www.cnblogs.com/grandyang/p/4843654.html</span>