三种方法求解两个数组的交集

Posted 2021-03-26 H.C

package com.Test;
 
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
 
//求两个数组的交集
public class FindIntersectionBetweenTwoArrays {
    
    //算法一：暴力搜索，时间复杂度O(n^2)，空间复杂度O(1)
    public ArrayList<Integer> getIntersection(int[] arr, int[] arr2) {
        
        ArrayList<Integer> res = new ArrayList<Integer>();
        if(arr == null || arr.length == 0 || arr2 == null || arr2.length == 0) {
            return res;
        }
        
        for(int i = 0;i < arr.length;i ++) {
            int temp = arr[i];
            for(int j = 0;j < arr.length;j ++) {
                if(arr2[j] == temp && !res.contains(temp)) {
                    res.add(temp);
                    break;
                }
            }
        }
        return res;
    }
    
    //算法二：先排序，然后定义两个指针，时间复杂度O(nlogn) (排序)，空间复杂度O(1)
    public ArrayList<Integer> getIntersection2(int[] arr, int[] arr2) {
        
        ArrayList<Integer> res = new ArrayList<Integer>();
        if(arr == null || arr.length == 0 || arr2 == null || arr2.length == 0) {
            return res;
        }
        
        Arrays.sort(arr);
        Arrays.sort(arr2);
        int i = 0;
        int j = 0;
        while(i < arr.length && j < arr2.length) {
            if(arr[i] < arr2[j]) {
                i ++;
            }
            else if(arr[i] > arr2[j]) {
                j ++;
            }
            else {
                if(!res.contains(arr[i])) {
                    res.add(arr[i]);
                }
                i ++;
                j ++;
            }
        }
        return res;
    }
    
    //算法三：map计数，时间复杂度O(n)，空间复杂度O(n)，空间换时间
    public ArrayList<Integer> getIntersection3(int[] arr, int[] arr2) {
        
        ArrayList<Integer> res = new ArrayList<Integer>();
        if(arr == null || arr.length == 0 || arr2 == null || arr2.length == 0) {
            return res;
        }
        
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for(int i = 0;i < arr.length;i ++) {
            int key = arr[i];
            if(map.containsKey(key)) {
                int value = map.get(key);
                value ++;
                map.put(key, value);
            }
            else {
                map.put(key, 1);
            }
        }
        for(int i = 0;i < arr2.length;i ++) {
            int key = arr2[i];
            if(map.containsKey(key) && !res.contains(key)) {
                res.add(key);
            }
        }
        return res;
        
    }
    
    public static void main(String[] args) {
        
        int[] arr = {1,2,5,3,3,4};
        int[] arr2 = {2,3,3,4,5,6};
        FindIntersectionBetweenTwoArrays fb = new FindIntersectionBetweenTwoArrays();
        List<Integer> res = fb.getIntersection3(arr, arr2);
        System.out.println(res);
    }
}
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版权声明：本文为CSDN博主「奇零可草」的原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接及本声明。
原文链接：https://blog.csdn.net/zhou15755387780/article/details/81317561