zoukankan      html  css  js  c++  java
  • [Algorithm] Find merge point of two linked list

    Assume we have two linked list, we want to find a point in each list, from which all the the nodes share the same value in both list. Then we call this point is "merge point".

     In the iamge, the merge point shoud be Node(7). 

      // Link A length 4, Link B length 5
      // All the node after the merge point should be the same
      // based on that, we can make sure that the merge point should
      // happens in range [x, c] or [y, c], it is not possible to happen at z
      // A      x - x - c - c
      // B  z - y - y - c - c
    function Node(val) {
      return {
        val,
        next: null
      };
    }
    
    function Link() {
      return {
        head: null,
        tail: null,
        length: 0,
        add(val) {
          const newNode = new Node(val);
    
          if (this.length === 0) {
            this.head = newNode;
            this.tail = newNode;
            this.tail.next = null;
          } else {
            let temp = this.tail;
            this.tail = newNode;
            temp.next = this.tail;
          }
    
          this.length++;
        }
      };
    }
    
     // O ( m + n ), Space: O(1)
    function findMergePoint(a, b) {
      // Link A length 4, Link B length 5
      // All the node after the merge point should be the same
      // based on that, we can make sure that the merge point should
      // happens in range [x, c] or [y, c], it is not possible to happen at z
      // A      x - x - c - c
      // B  z - y - y - c - c
      let diff, headA = a.head, headB = b.head;
      if (a.length > b.length) { // O(1)
        diff = a.length - b.length; 
        [a, b] = [b, a];
      } else {
        diff = b.length - a.length;
      }
    
      // reach +diff step for longer 
      for (let i = 0; i < diff; i++) {
        headB = headB.next; // O(m)
      }
    
      while (headA != null && headB.val != null ) { // O(n)
        if (headA.val === headB.val) {
          return headA;
        }
    
        headA = headA.next;
        headB = headB.next;
      } 
    
      return null;
    }
    
    const lA = new Link();
    lA.add(4);
    lA.add(6);
    lA.add(7);
    lA.add(1);
    
    const lB = new Link();
    lB.add(9);
    lB.add(3);
    lB.add(5);
    lB.add(7);
    lB.add(1);
    
    console.log(findMergePoint(lA, lB)); // Object {val: 7, next: Object}
  • 相关阅读:
    C++内存泄露的有效预防方法:谁使用,谁删除 (1.2)
    工作日志2014-08-28
    【2012.1.24更新】不要再在网上搜索eclipse的汉化包了!
    关于ActionContext.getContext()的使用方法心得
    Android开发(20)--RadioGroup的使用
    站点防止攻击
    小强的HTML5移动开发之路(50)——jquerymobile页面初始化过程
    我是怎样成长为系统架构师的
    辛星站点架构师笔记第四篇
    strcpy_s与strcpy的比較
  • 原文地址:https://www.cnblogs.com/Answer1215/p/10650273.html
Copyright © 2011-2022 走看看