LintCode 548. Intersection of Two Arrays II

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本文介绍了一种使用unordered_map实现的数组交集算法，该算法能够有效地找出两个整数数组的交集元素及其出现次数，并提供了代码示例。讨论了不同情况下算法的优化策略。

Intersection of Two Arrays II
中文English
Given two arrays, write a function to compute their intersection.

Example
Example1

Input:
nums1 = [1, 2, 2, 1], nums2 = [2, 2]
Output:
[2, 2]
Example2

Input:
nums1 = [1, 1, 2], nums2 = [1]
Output:
[1]
Challenge
What if the given array is already sorted? How would you optimize your algorithm?
What if nums1’s size is small compared to num2’s size? Which algorithm is better?
What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?
Notice
Each element in the result should appear as many times as it shows in both arrays.
The result can be in any order.

解法1：用unordered_map。
注意：

引用也可以指向vector。这样就可以搞定nums1和nums2大小不定的问题。

代码如下:

class Solution {
public:
    /**
     * @param nums1: an integer array
     * @param nums2: an integer array
     * @return: an integer array
     */
    vector<int> intersection(vector<int> &nums1, vector<int> &nums2) {
        int M = nums1.size();
        int N = nums2.size();
        
        if (M == 0 || N == 0) return vector<int>();
        
        vector<int> & large = (M >= N) ? nums1 : nums2;
        vector<int> & small = (M < N) ? nums1 : nums2;
    
        unordered_map<int, int> umLarge; //(number, freq)
        
        for (int i = 0; i < large.size(); ++i) {
            if (umLarge.find(large[i]) == umLarge.end()) {
                umLarge[large[i]] = 1;
            } else {
                umLarge[large[i]]++;
            }
        }

        vector<int> result;
        
        for (int i = 0; i < small.size(); ++i) {
            if (umLarge.find(small[i]) != umLarge.end() &&
                umLarge[small[i]] > 0) {
                    result.push_back(small[i]);
                    umLarge[small[i]]--;
                }   
        }
        
        return result;    
    }

};