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  • leetcode Intersection of Two Linked Lists

    Write a program to find the node at which the intersection of two singly linked lists begins.

    For example, the following two linked lists:

    A:          a1 → a2
                       ↘
                         c1 → c2 → c3
                       ↗            
    B:     b1 → b2 → b3
    

    begin to intersect at node c1.

    Notes:

      • If the two linked lists have no intersection at all, return null.
      • The linked lists must retain their original structure after the function returns.
      • You may assume there are no cycles anywhere in the entire linked structure.
      • Your code should preferably run in O(n) time and use only O(1) memory
      • 从后对齐,只需比较后面的元素,有没有相同的
      • 没有中间有一个相同的,后面又不同的情况
      • /**
         * 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) {
                ListNode result = null;
        		ListNode tempa=headA;
        		ListNode tempb=headB;
        		if (headA == null || headB == null) {
        			return null;
        		}
        		int a = 1, b = 1;
        		while (headA.next != null) {
        			a++;
        			headA = headA.next;
        		}
        		while (headB.next != null) {
        			b++;
        			headB = headB.next;
        		}
        		int temp = 0;
        		if (a >= b) {
        			temp = a - b;
        			while (temp != 0) {
        				tempa = tempa.next;
        				temp--;
        			}
        		} else {
        			temp = b - a;
        			while (temp != 0) {
        				tempb = tempb.next;
        				temp--;
        			}
        		}
        		while (tempa!= null) {
        			if (tempa == tempb) {
        				result = tempa;
        				break;
        			} else {
        				tempa = tempa.next;
        				tempb = tempb.next;
        			}
        		}
        		return result;
            }
        }
        

          

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  • 原文地址:https://www.cnblogs.com/lilyfindjobs/p/4222971.html
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