zoukankan      html  css  js  c++  java
  • [Algorithm] 350. Intersection of Two Arrays II

    Given two arrays, write a function to compute their intersection.

    Example 1:

    Input: nums1 = [1,2,2,1], nums2 = [2,2]
    Output: [2,2]
    

    Example 2:

    Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
    Output: [4,9]

    Note:

    • Each element in the result should appear as many times as it shows in both arrays.
    • The result can be in any order.

    Not good enough approach

    var intersect = function(nums1, nums2) {
        let r = [];
        // get larger array
        const len1 = nums1.length;
        const len2 = nums2.length;
        
        // larger & smaller
        const larger = len1 > len2 ? nums1: nums2;
        const smaller = len1 > len2 ? nums2: nums1;
        
        // conver larger array to object
        let hashed = {};
        for (let n of larger) {
            if (n in hashed) {
                hashed[n]++;
            } else {
                hashed[n] = 1;
            }
        }
     
        // loop over smaller array
        for (let n of smaller) {
            if (`${n}` in hashed) {
                r.push(n);
                hashed[n] = hashed[n]-1;
                if (hashed[n] === 0) {
                    delete hashed[n];
                }
            }
        }
        
        return r;
    };

    The reason that code above is not good enough is because,

    1. we use larger array as lookup, this cause more memory usage. -- actually we need to use smaller array as lookup

    2. we use 'len1, len2, samller, larger' extra storage, we can actully swap nums1 and nums by one extra function call.

    var intersect = function(nums1, nums2) {
    
        if (nums1.length > nums2.length) {
            return intersect(nums2, nums1);
        }
        
        // conver samller array to object
        let hashed = {};
        for (let n of nums2) {
            if (n in hashed) {
                hashed[n]++;
            } else {
                hashed[n] = 1;
            }
        }
        
        let r = [];
        // loop over smaller array
        for (let n of nums1) {
            if (hashed[n] > 0) {
                r.push(n);
                hashed[n] = hashed[n]-1;
            }
        }
        
        return r;
    }

    Follow up:

    • What if the given array is already sorted? How would you optimize your algorithm?
    • What if nums1's size is small compared to nums2's size? Which algorithm is better?
    • What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

    Will write another post for the follow up questions.

  • 相关阅读:
    Vmware Vsphere WebService之vijava 开发一-vcenter连接、及集群信息获取
    Vmware Vsphere WebService SDK开发(第一讲)-基本知识学习
    RabbitMQ安装以及java使用(二)
    redis单机安装以及集群搭建(redis-6.2.6)
    Spring Cloud Gateway中Filter获取Request Body的几种方式
    电子发票插入微信卡包之PDF上传
    Elasticsearch集群搭建详解
    微服务的设计原则
    centos 7.4 64位 mysql的安装
    RabbitMQ安装以及java使用(一)
  • 原文地址:https://www.cnblogs.com/Answer1215/p/12040497.html
Copyright © 2011-2022 走看看