Intersection of Two Arrays II

最新推荐文章于 2024-03-31 22:07:26 发布

转载最新推荐文章于 2024-03-31 22:07:26 发布 · 52 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/FLAGyuri/p/5582719.html

本文介绍了一种使用双指针及排序的方法来计算两个数组的交集，并确保每个元素出现次数符合输入数组中的频率。该算法适用于需要计算重复元素交集的问题。

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

Note:

Each element in the result should appear as many times as it shows in both arrays.

The result can be in any order.

解题思路：双指针，排序

 1 public class Solution {
 2     public int[] intersect(int[] nums1, int[] nums2) {
 3         Arrays.sort(nums1);
 4         Arrays.sort(nums2);
 5         int i = 0, j = 0;
 6         int index = 0;
 7         int[] temp = new int[nums1.length];
 8         while (i < nums1.length && j < nums2.length) {
 9             if (nums1[i] == nums2[j]) {
10                 temp[index++] = nums1[i];
11                 ++i;
12                 ++j;
13             } else if (nums1[i] < nums2[j]) {
14                 i++;
15             } else {
16                 j++;
17             }
18         }
19         int[] result = new int[index];
20         for (int k = 0; k < index; k++) {
21             result[k] = temp[k];
22         }
23         return result;
24     }
25 }