Find the Duplicate Number问题及解法

问题描述:

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.

问题分析：

本题看似附加要求很多，实质上跟142. Linked List Cycle II 的解法类似，利用双指针，找到fast和slow相遇的值，再去找到circle的入口即可。

过程详见代码：

class Solution {public:    int findDuplicate(vector<int>& nums) {        int slow = nums[0],fast = nums[nums[0]];while (slow != fast){slow = nums[slow];fast = nums[nums[fast]];}fast = 0;while (slow != fast){slow = nums[slow];fast = nums[fast];}return slow;    }};