LeetCode 350 Intersection of Two Arrays II

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。

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> res;
        int len1 = nums1.size(), len2 = nums2.size();
        
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        
        int i = 0, j = 0;
        while (i < len1 && j < len2) {
            if (nums1[i] < nums2[j]) {
                i++;
            }
            else if (nums1[i] > nums2[j]) {
                j++;
            }
            else {
                res.push_back(nums1[i]);
                i++;
                j++;
            }
        }
        
        return res;
    }
};

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。

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> res;
        map<int, int> m;
        int len1 = nums1.size(), len2 = nums2.size();
        
        for (int i = 0; i < len1; i++) {
            m[nums1[i]]++;
        }
        
        for (int i = 0; i < len2; i++) {
            if (m[nums2[i]]-- > 0) {
                res.push_back(nums2[i]);
            }
        }
        
        return res;
    }
};