数组和矩阵（1）——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:

1. You must not modify the array (assume the array is read only).  //不能排序
2. You must use only constant, O(1) extra space.                         // 不能用哈希表
3. Your runtime complexity should be less than O(n²).                  //不能暴力求解
4. There is only one duplicate number in the array, but it could be repeated more than once.

https://segmentfault.com/a/1190000003817671#articleHeader4

考虑：

1. 暴力求解，选择一个数，看有没有重复的；
2. 哈希表
3. 排序后遍历
4. 二分法
5. 设置快慢指针，映射找环法

public class Solution {
    public int findDuplicate(int[] nums) { //映射找环
        int n = nums.length - 1;
        int pre = 0;
        int last = 0;
        do {
            pre = nums[pre];
            last = nums[nums[last]];
        } while(nums[pre] != nums[last]);
        last = 0;
        while(nums[pre] != nums[last]) {
            pre = nums[pre];
            last = nums[last];
        }
        return nums[last];
    }
}