linked-list-cycle-ii-LeetCode

linked-list-cycle-ii

题目描述

Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull. Follow up: Can you solve it without using extra space?

思路

1. 设置两个指针nodeFast和nodeSlow，nodeSlow每次走一步，nodeFast每次走两步，如果存在环，则没有空指针，而且两个节点相遇的时候nodeFast比nodeSlow多走的节点数正好是环里包括的节点数
2. 然后将nodeFast置为head从头走，nodeSlow不变，这次两个节点每次都走一步，也就是说nodeSlow比nodeFast一直多走一个环，当他们相遇时，刚好是在环的起点处，然后就再也不分开了

代码

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
        if (head == null) {
			return null;
		}
		
		ListNode nodeFast = head;
		ListNode nodeSlow = head;
		do {
			if (nodeFast.next == null) {
				return null;
			}
			nodeFast = nodeFast.next;
			nodeSlow = nodeSlow.next;
			if (nodeFast.next == null) {
				return null;
			}
			nodeFast = nodeFast.next;
		} while (nodeFast != nodeSlow);
		
		nodeFast = head;
		while (nodeFast != nodeSlow) {
			nodeFast = nodeFast.next;
			nodeSlow = nodeSlow.next;
		}
		return nodeSlow;
    }
}

linked-list-cycle-ii-LeetCode

原文：http://www.cnblogs.com/rosending/p/5770123.html