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:
    小老虎CSDN博客流量分析
    C#中字符串的内存分配与驻留池
    最简单的基于FFmpeg的封装格式处理:视音频分离器(demuxer)
    HDU 3037 Saving Beans(Lucas定理的直接应用)
    Linux中IRC通讯工具Pidgin的基本用法
    jQuery整理笔记八----jQuery的Ajax
    Android ServiceManager启动
    C++开源码项目汇总
    虚拟现实游戏的十大误区
    适用android的MVP:怎样组织展示层
  • 原文地址:https://www.cnblogs.com/Answer1215/p/10650273.html
Copyright © 2011-2022 走看看