mysql+两个数组交集_三种方法求解两个数组的交集

packagecom.Test;importjava.util.ArrayList;importjava.util.Arrays;importjava.util.HashMap;importjava.util.List;importjava.util.Map;//求两个数组的交集

public classFindIntersectionBetweenTwoArrays {//算法一：暴力搜索，时间复杂度O(n^2)，空间复杂度O(1)

public ArrayList getIntersection(int[] arr, int[] arr2) {

ArrayList res = new ArrayList();if(arr == null || arr.length == 0 || arr2 == null || arr2.length == 0) {returnres;

}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;

}

}

}returnres;

}//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 getIntersection2(int[] arr, int[] arr2) {

ArrayList res = new ArrayList();if(arr == null || arr.length == 0 || arr2 == null || arr2.length == 0) {returnres;

}

Arrays.sort(arr);

Arrays.sort(arr2);int i = 0;int j = 0;while(i < arr.length && j

i++;

}else if(arr[i] >arr2[j]) {

j++;

}else{if(!res.contains(arr[i])) {

res.add(arr[i]);

}

i++;

j++;

}

}returnres;

}//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 getIntersection3(int[] arr, int[] arr2) {

ArrayList res = new ArrayList();if(arr == null || arr.length == 0 || arr2 == null || arr2.length == 0) {returnres;

}

Map map = new HashMap();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);

}

}returnres;

}public static voidmain(String[] args) {int[] arr = {1,2,5,3,3,4};int[] arr2 = {2,3,3,4,5,6};

FindIntersectionBetweenTwoArrays fb= newFindIntersectionBetweenTwoArrays();

List res =fb.getIntersection3(arr, arr2);

System.out.println(res);

}

}

————————————————

版权声明：本文为CSDN博主「奇零可草」的原创文章，遵循 CC4.0 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