【LeeCode】349. 两个数组的交集

【题目描述】

给定两个数组 ​​nums1​​ 和 ​​nums2​​ ，返回 它们的交集 唯一 的。我们可以 不考虑输出结果的顺序 。

​​​​https://leetcode.cn/problems/intersection-of-two-arrays/​​

【示例】

【代码】admin

思路：2个循环

package com.company;
// 2023-03-09
import java.util.*;

class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < nums1.length; i++){
            for (int j = 0; j < nums2.length; j++){
                if (nums1[i] == nums2[j] && !list.contains(nums2[j])){
                    list.add(nums2[j]);
                }
            }
        }
        int[] res = new int[list.size()];
        for (int i = 0; i < list.size(); i++){
            res[i] = list.get(i);
        }
        return res;
    }
}

public class Test {
    public static void main(String[] args) {
        new Solution().intersection(new int[]{1,2,2,1}, new int[]{2, 2});   // 输出：[2]
        new Solution().intersection(new int[]{4,9,5}, new int[]{9,4,9,8,4});   // 输出：[9,4]
    }
}

【代码】admin

思路： 基于内部set取交集

package com.company;
// 2023-03-09
import java.util.*;
import java.util.stream.Collectors;

class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {

        Set<Integer> set1 = Arrays.stream(nums1).boxed().collect(Collectors.toSet());
        Set<Integer> set2 = Arrays.stream(nums2).boxed().collect(Collectors.toSet());
        set1.retainAll(set2);
        int[] res = new int[set1.size()];
        int index = 0;
        for (Integer x : set1){
            res[index] = x;
            index++;
        }
        System.out.println(Arrays.toString(res));
        return res;
    }
}

public class Test {
    public static void main(String[] args) {
        new Solution().intersection(new int[]{1,2,2,1}, new int[]{2, 2});   // 输出：[2]
        new Solution().intersection(new int[]{4,9,5}, new int[]{9,4,9,8,4});   // 输出：[9,4]
    }
}

【代码】admin

思路： 把一个加入hash中然后判断

package com.company;
// 2023-03-09
import java.util.*;
import java.util.stream.Collectors;

class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        List<Integer> list = Arrays.stream(nums2).boxed().collect(Collectors.toList());
        List<Integer> res = new ArrayList<>();
        for (int i = 0; i < nums1.length; i++){
            if (list.contains(nums1[i]) && !res.contains(nums1[i])){
                res.add(nums1[i]);
            }
        }
        int[] re = new int[res.size()];
        for (int i = 0; i < res.size(); i++){
            re[i] = res.get(i);
        }
        return re;
    }
}

public class Test {
    public static void main(String[] args) {
        new Solution().intersection(new int[]{1,2,2,1}, new int[]{2, 2});   // 输出：[2]
        new Solution().intersection(new int[]{4,9,5}, new int[]{9,4,9,8,4});   // 输出：[9,4]
    }
}