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

不能排序，因为只能用O(1)的space，也不能hash，所以用环来接解决。
先用快慢指针找出环的起点，再用两个指针找出相同的值。

代码

class Solution {
    public int findDuplicate(int[] nums) {
        int slow = nums[0];
        int fast = nums[0];
        while(true){           
            slow = nums[slow];
            fast = nums[nums[fast]];
            if(fast == slow) break;
        }
        int p1 = fast;
        int p2 = nums[0];
        while(p1!=p2){
            p1 = nums[p1];
            p2 = nums[p2];
        }
        return p1;        
    }
}