zoukankan      html  css  js  c++  java
  • 0350. Intersection of Two Arrays II (E)

    Intersection of Two Arrays II (E)

    题目

    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.

    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?

    题意

    取两数组的交集,包含所有重复元素

    思路

    普通做法直接用HashMap处理即可。

    排序情况下用类似于归并排序的方法处理。


    代码实现

    Java

    Hash

    class Solution {
        public int[] intersect(int[] nums1, int[] nums2) {
            List<Integer> list = new ArrayList<>();
            Map<Integer, Integer> record = new HashMap<>();
            for (int num : nums1) {
                record.putIfAbsent(num, 0);
                record.put(num, record.get(num) + 1);
            }
    
            for (int num : nums2) {
                if (record.containsKey(num) && record.get(num) > 0) {
                    list.add(num);
                    record.put(num, record.get(num) - 1);
                }
            }
    
            int[] res = new int[list.size()];
            int i = 0;
            for (int num : list) {
                res[i++] = num;
            }
    
            return res;
        }
    }
    

    排序

    class Solution {
        public int[] intersect(int[] nums1, int[] nums2) {
            List<Integer> list = new ArrayList<>();
            Arrays.sort(nums1);
            Arrays.sort(nums2);
            int p1 = 0, p2 = 0;
            while (p1 < nums1.length && p2 < nums2.length) {
                if (nums1[p1] > nums2[p2]) {
                    p2++;
                } else if (nums1[p1] < nums2[p2]) {
                    p1++;
                } else {
                    list.add(nums1[p1]);
                    p1++;
                    p2++;
                }
            }
    
            int[] res = new int[list.size()];
            int i = 0;
            for (int num : list) {
                res[i++] = num;
            }
    
            return res;
        }
    }
    
  • 相关阅读:
    从例图中学习思维导图的基本概念
    名企面试官精讲典型编程题之C++篇
    PPT,要你好看(全彩)
    文字即艺术
    MindManager中读图工具的使用
    思维导图的三招十八式
    放之四海皆适用的设计原则(一)
    MindManager的例图资源
    内容营销11金规
    专业嵌入式软件开发——全面走向高质高效编程(含DVD光盘1张)
  • 原文地址:https://www.cnblogs.com/mapoos/p/13211469.html
Copyright © 2011-2022 走看看