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

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:

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：

1）理论证明：鸽笼原理，说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。用反证法证明：假设每个容器里面的东西至多有一个，那么m个容器中的东西至多有m < n个，与一共有n个东西相矛盾，所以假设不成立，鸽笼原理成立。

2）代码实现：由于不允许对数组进行排序，也不允许使用额外的空间，所以少于O(n^2)的时间复杂度的解法的唯一可能性就是二分查找。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：

class Solution {
public:
int findDuplicate(vector<int>& nums) {
int len = nums.size();
int left = 1, right = len;
while(left <= right) {
int mid = left + (right - left) / 2;
int 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;
}
};