linked-list-cycle-ii
2016-07-19 17:06
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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?
解析:
比I麻烦点的就是要找到循环开始点。I只是判断是否循环。画个图(很丑勿喷)
假设在红色凸起的地方相遇了。
F走的路程应该是S的两倍
S = x + y
F = x + y + z + y = x + 2y + z
2*S = F
2x+2y = x + 2y + z
得到x = z
也就是从head到环开始的路程 = 从相遇到环开始的路程
那么。。。只要S和F相遇了,我们拿一个从头开始走,一个从相遇的地方开始走
两个都走一步,那么再次相遇必定是环的开始节点!
package com.leetcode;
public class HasCycle2 {
public ListNode hasCycle(ListNode head) {
ListNode slow = head;
ListNode fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast)
break;
}
if (fast == null || fast.next == null)
return null;
slow = head;
while (slow != fast) {
slow = slow.next;
fast = fast.next;
}
return fast;
}
}
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