弗洛伊德判圈算法

先来看链表是否有环的判断设链表起始节点为H，回路起始节点为S，两指针第一次相遇节点为M。设回路的长度为c，S到M的距离为c1，M到S的距离为c2。设H到S的距离为d。设hare每次移动2，tortoise每次移动1。移动k次后两者相遇。不难发现，H到M的距离为k。为了便于理解，请参考下图：由已知条件，可得到下列等式：hare和tortoise相遇时，hare套了tortoise若干圈。则有：2k – k = nc = k，n为正整数。从上图中定义的距离关系，可以得到c = c1 + c2k = d + c1由以上格式变换可得： d = nc – c1 = c2 + (n-1)c该式具有什么重要意义呢？假设2个指针分别从H和M出发，都是每次移动1的距离，他们会在哪里相遇？一个指针从H出发移动d后到达M。另一指针移动 c2 + (n-1)c，(n-1)c意味着绕了n-1圈又回到M，实际上相当于走了c2的距离，到达M。我们发现，两指针必然在S相遇。

代码如下：

 public ListNode detectCycle(ListNode head) {        if (head == null) return null;        ListNode fast = head, slow = head;        while (fast.next != null && fast.next.next != null) {            slow = slow.next;            fast = fast.next.next;            if (slow == fast) {                fast = head;                while (fast!=slow){                    fast=fast.next;                    slow = slow.next;                }                return fast;            }                    }        return null;    }

再看一个问题

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).
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.

这个问题的限制条件比较多。不能修改数组元素，只能用常量空间，复杂度< O(n2).用鸽笼原理很好证明必有重复元素。把元素取值1到n看成n个笼子，n+1个数放到n个笼子必有重复。

举个例子：

4,1,2,3,1

重复的元素构成环

1,2,3,4,5

4,1,2,3,1

1->4, 4->3,3->2,2->1所以1是重复的

     int next(int array[], int num) {         return array[num];    }    public int findDuplicate(int[] nums) {        int fast = nums[0], slow = nums[0];        while (true) {            slow = next(nums, slow);            fast = next(nums, next(nums, fast));            if (slow == fast) {                fast = nums[0];                while (fast != slow) {                    fast = next(nums, fast);                    slow = next(nums, slow);                }                return fast;            }        }            }