350. Intersection of Two Arrays II

Total Accepted: 3237  Total Submissions: 7617  Difficulty: Easy

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.

Follow up:

• 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?

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  (E) Intersection of Two Arrays

分析：

先让数组有序，再双指针遍历即可，有相同元素就获取结果，然后判断下一对

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> result;
        sort(nums1.begin(),nums1.end());
        sort(nums2.begin(),nums2.end());//排序
        int i=0;
        int j=0;
        while(i < nums1.size() && j < nums2.size())//双指针遍历
        {
            if(nums1[i] < nums2[j])
                i++;
            else if(nums1[i] > nums2[j])//移动
                j++;
            else//相等，获取结果
             {
                result.push_back(nums1[i]);
                i++;j++;
             }
        }
        return result;  
    }
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/51488374

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895