350. Intersection of Two Arrays II

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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 nums2‘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?

===================

返回交叉元素,

注意:结果中每个元素应该出现和两个数组相同的次数,结果的顺序任意.

思路:排序+遍历

==

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> re;
        sort(nums1.begin(),nums1.end());
        sort(nums2.begin(),nums2.end());

        vector<int>::iterator it1 = nums1.begin();
        vector<int>::iterator it2 = nums2.begin();
        for(;it1!=nums1.end() && it2!=nums2.end();){
            if(*it1<*it2) it1++;
            else if(*it1 > *it2) it2++;
            else{
                re.push_back(*it1);
                it1++;
                it2++;
            }
        }
        return re;
    }
};

350. Intersection of Two Arrays II

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原文地址：http://www.cnblogs.com/li-daphne/p/5607863.html