leetcode [Intersection of Two Linked Lists]//待整理多种解法
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/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */public class Solution { public ListNode getIntersectionNode(ListNode headA, ListNode headB) { //解法一:时间是O(Max(m,n)),但是辅助空间是O(Max(m,n)),而不是O(1) //既然要找两个列表共用的第一个结点,那么就从两个列表的最后一个结点开始比较,不能反序(因为题目要求原来的两个列表的结构不能改变),那么就用两个栈来存储 ListNode res = null; ListNode shiftA = headA; ListNode shiftB = headB; Stack<ListNode> stackA = new Stack<>(); Stack<ListNode> stackB = new Stack<>(); while(shiftA != null){ stackA.push(shiftA); shiftA = shiftA.next; } while(shiftB != null){ stackB.push(shiftB); shiftB = shiftB.next; } while(!stackA.isEmpty() && !stackB.isEmpty()){ shiftA = stackA.pop(); shiftB = stackB.pop(); if(shiftA == shiftB){ res = shiftA; } } return res; }}
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