LeetCode-Find the Duplicate Number

Description：
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.

Example 1:

Input: [1,3,4,2,2]
Output: 2

Example 2:

Input: [3,1,3,4,2]
Output: 3

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.

题意：找出数组中重复出现得元素；并且时间复杂度不超过O(n²)，空间复杂度为O(1)；

解法：这里用到了Floyd判圈算法(龟兔赛跑算法)；第一步找出数组中存在的环(数组元素代表的就是下一个元素的位置，因为这里元素的大小在[1,n])，第二步找到环的入口，如图所示：

Java

class Solution {
    public int findDuplicate(int[] nums) {
        int tortoise = nums[0];
        int hare = nums[0];
        do {
            tortoise = nums[tortoise];
            hare = nums[nums[hare]];
        } while (tortoise != hare);
        int ptr = nums[0];
        while (ptr != tortoise) {
            ptr = nums[ptr];
            tortoise = nums[tortoise];
        }
        return ptr;
    }
}

参考文献：
[1]：https://blog.****.net/xyzxiaoxiong/article/details/78761940
[2]：https://en.wikipedia.org/wiki/Cycle_detection#Floyd’s_Tortoise_and_Hare