算法系列（5） leetcode350

最新推荐文章于 2022-10-10 16:52:52 发布

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最新推荐文章于 2022-10-10 16:52:52 发布

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分类专栏：算法文章标签： JavaScript 算法 leetcode

本文链接：https://blog.youkuaiyun.com/u014032819/article/details/78384413

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本文介绍了一个用于计算两个数组交集的算法实现。该算法通过双重循环遍历两个数组，找到相同的元素并将其添加到结果数组中。同时，文章探讨了在不同情况下（如已排序数组、大小不等的数组等）如何优化此算法。

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

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

Note:

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

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?

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersect = function(nums1, nums2) {
    var a = [];
    for (var j = 0; j<nums1.length; j++) {
        for (var i = 0; i < nums2.length; i++) {
            //console.log(w);
            if (nums1[j] == nums2[i]) {
                a.push(nums1[j]);
                nums1.splice(j,1);
                nums2.splice(i, 1);
                i--;
                j--;
                // console.log(nums2);
            }
        }
    }

    return a;
};