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package leetcode_1To300;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
/**
* 本代码来自 Cspiration,由 @Cspiration 提供
* 题目来源:http://leetcode.com
* - Cspiration 致力于在 CS 领域内帮助中国人找到工作,让更多海外国人受益
* - 现有课程:Leetcode Java 版本视频讲解(1-900题)(上)(中)(下)三部
* - 算法基础知识(上)(下)两部;题型技巧讲解(上)(下)两部
* - 节省刷题时间,效率提高2-3倍,初学者轻松一天10题,入门者轻松一天20题
* - 讲师:Edward Shi
* - 官方网站:https://cspiration.com
* - 版权所有,转发请注明出处
*/
public class _350_IntersectionofTwoArraysII {
/**
* 350. Intersection of Two Arrays II
* Given two arrays, write a function to compute their intersection.
Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].
* @param nums1
* @param nums2
* @return
*/
// HashMap, time : O(n), space : O(n);
public static int[] intersect(int[] nums1, int[] nums2) {
HashMap<Integer, Integer> map = new HashMap<>();
List<Integer> ret = new ArrayList<>();
for (int i = 0; i < nums1.length; i++) {
if (map.containsKey(nums1[i])) {
map.put(nums1[i],map.get(nums1[i]) + 1);
} else {
map.put(nums1[i], 1);
}
}
for (int i = 0; i < nums2.length; i++) {
if (map.containsKey(nums2[i])) {
if (map.get(nums2[i]) > 0) {
ret.add(nums2[i]);
map.put(nums2[i], map.get(nums2[i]) - 1);
}
}
}
int[] res = new int[ret.size()];
int k = 0;
for (Integer num : ret) {
res[k++] = num;
}
return res;
}
// Arrays.sort time : O(nlogn) space : O(n);
public static int[] intersect2(int[] nums1, int[] nums2) {
Arrays.sort(nums1);
Arrays.sort(nums2);
List<Integer> ret = new ArrayList<>();
int i = 0;
int j = 0;
while (i < nums1.length && j < nums2.length) {
if (nums1[i] < nums2[j]) {
i++;
} else if (nums1[i] > nums2[j]) {
j++;
} else {
ret.add(nums1[i]);
i++;
j++;
}
}
int[] res = new int[ret.size()];
int k = 0;
for (Integer num : ret) {
res[k++] = num;
}
return res;
}
}