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LeetCode 350 Intersection of Two Arrays II

2024-08-18 05:00:02   222人阅读

Problem:

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

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

  • Each element in the result should appear as many times as it shows in both arrays.
  • The result can be in any order.

Summary:

求两集合交集，允许有重复元素出现。

Analysis:

1. sort + merge

    首先将两个数组从小到大排序，后分别用两个指针指向两个数组开头比较大小，将小的数组指针后移。直至两指针指向数字相等时考虑将数字放入res中。

　 这道题不用考虑res中是否包含该数字。

 1 class Solution {
 2 public:
 3     vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
 4         vector<int> res;
 5         int len1 = nums1.size(), len2 = nums2.size();
 6         
 7         sort(nums1.begin(), nums1.end());
 8         sort(nums2.begin(), nums2.end());
 9         
10         int i = 0, j = 0;
11         while (i < len1 && j < len2) {
12             if (nums1[i] < nums2[j]) {
13                 i++;
14             }
15             else if (nums1[i] > nums2[j]) {
16                 j++;
17             }
18             else {
19                 if (res.empty() || res.back() != nums1[i]) {
20                     res.push_back(nums1[i]);
21                 }
22                 
23                 i++;
24                 j++;
25             }
26         }
27         
28         return res;
29     }
30 };

2. Hash表建立数组中数字和出现次数的映射，再在第二个数组中进行查找，注意每找到一个，要在map中将该元素出现次数减1。

 1 class Solution {
 2 public:
 3     vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
 4         vector<int> res;
 5         map<int, int> m;
 6         int len1 = nums1.size(), len2 = nums2.size();
 7         
 8         for (int i = 0; i < len1; i++) {
 9             m[nums1[i]]++;
10         }
11         
12         for (int i = 0; i < len2; i++) {
13             if (m[nums2[i]]-- > 0) {
14                 res.push_back(nums2[i]);
15             }
16         }
17         
18         return res;
19     }
20 };

LeetCode 350 Intersection of Two Arrays II


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