【LeetCode從零單排】No.160 Intersection of Two Linked Lists

李博Garvin發表於2015-03-10
原文網址 : https://blog.csdn.net/buptgshengod/article/details/44174967
LeetCode

題目

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.

Credits:
Special thanks to @stellari for adding this problem and creating all test cases.


題目要求取兩個連結串列的交點，而且時間複雜度必須是O(n)，所以就不能用巢狀迴圈的方法。用瞭如下方法，先計算兩個連結串列的各自長度，將長連結串列節點向下移動兩連結串列長度差，再計算。


程式碼

/**
 * 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) {
        if(headA==null || headB==null) return null;
      
        int Aindex_length=getLength(headA);
        int Bindex_length=getLength(headB);
        int dis=Math.abs(Aindex_length-Bindex_length);
        
        
            
        
        ListNode Aindex=headA;
        ListNode Bindex=headB;
        if(Aindex_length>=Bindex_length){
          for(int i=0;i<dis;i++){
              Aindex=Aindex.next;
          }  
          while(Bindex!=null){
              if(Aindex.val==Bindex.val){return Aindex;}
              else{
                  Aindex=Aindex.next;
                  Bindex=Bindex.next;
                 }
          }
        }
       
         Aindex=headA;
         Bindex=headB;
      if(Aindex_length<Bindex_length){
          for(int i=0;i<dis;i++){
              Bindex=Bindex.next;
          }  
          while(Aindex!=null){
              if(Aindex.val==Bindex.val){return Aindex;}
              else{
                  Aindex=Aindex.next;
                  Bindex=Bindex.next;
                 }
          }
        }
        
        
        return null;
    }
    public int getLength(ListNode head){
        ListNode index=head;
        int length=1;
        while(index.next!=null){
            index=index.next;
            length++;
        }
        return length;
    }
}




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* 本文來自部落格  “李博Garvin“

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