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:
    JavaScript----数组方法
    JavaScript----数组
    JavaScript----Array.foreach()
    JavaScript----数字及数字方法
    JavaScript----函数,对象及字符串方法
    设计模式@第5章:单例设计模式
    设计模式@第4章:设计模式概述
    设计模式@第3章:UML 类图
    部署方案@常用软件的安装
    应用框架@SpringBoot
  • 原文地址:https://www.cnblogs.com/Answer1215/p/12040497.html
Copyright © 2011-2022 走看看