https://leetcode.com/problems/intersection-of-two-arrays/?tab=Description 一开始没弄清题意,以为是像intersection of two linked lists那样找交点,但实际上不是。其实是找两个数组的交集,即把两个数组中都有的元素不重不漏地找出来,所以其实比linked list的交点要简单。 思路1:Hash Set 首先把一个数组的所...
Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order. Example 1: Input: nums1 = [1,2,2,1], nums2 = [2,2] Output: [2,2] ...
What if the given array is already sorted? How would you optimize your algorithm? What ifnums1's size is small compared tonums2's size? Which algorithm is better? What if elements ofnums2are stored on disk, and the memory is limited such that you cannot load all elements into the memo...
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 ...
2、The result can be in any order. 要完成的函数: vector<int> intersection(vector<int>& nums1, vector<int>& nums2) 说明: 1、给定两个vector,求其交集。交集是一个集合,所以不能出现重复元素,集合不要求顺序。 2、最直接的思路是模仿人类思维,设置两个set,分别存储nums1和nums2的数值。set中没有...
LeetCode.349--intersection-of-two-arrays 一、题目链接 两个数组的交集 二、题目描述 给定两个数组nums1和nums2,返回它们的交集。输出结果中的每个元素一定是 唯一 的。我们可以 不考虑输出结果的顺序 。
https://leetcode.com/explore/interview/card/top-interview-questions-easy/92/array/674/leetcode.com/explore/interview/card/top-interview-questions-easy/92/array/674/ Solutions: 1. Sort and merge: We firstly sort the two arrays and use two pointers to compare the elements in the two arrays...
Consider the set S = {2, 3, 4}. For each interval, there are at least 2 elements from S in the interval. Also, there isn't a smaller size set that fulfills the above condition. Thus, we output the size of this set, which is 3. ...
* @return an integer array */publicint[]intersection(int[]nums1,int[]nums2){// Write your code hereArrays.sort(nums1);Arrays.sort(nums2);inti=0,j=0;int[]temp=newint[nums1.length];intindex=0;while(i<nums1.length&&j<nums2.length){if(nums1[i]==nums2[j]){if(index==0||temp...
//leetcode.com/discuss/interview-question/914249/Uber-or-Phone-or-Union-and-Intersection-of-Two-Sorted-Interval-Lists */ public class UnionAndIntersectionOfTwoSortedIntervalLists { public static void main(String[] args) { } public int[][] intersectionOfTwoIntervals(int[][] a, int[][] b) ...