Description

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

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

Note:
Each element in the result must be unique.
The result can be in any order.

My solution

自然的想法是把nums1的每个数和nums2的每个数对比,复杂度为O(nm).如果排好序的呢? 那就是nlgn和mlgm,一般情况下这个方式更小一点.
考虑到有很多重复数字, 并且问题的根本在于判断"共有元素",所以很容易联想到python方式, 把数组专为set,用 in / not in判断.没看过源码,也不知道这个过程复杂度是怎样的 ==
下为先排序的方式:

class Solution {
public:vector<int> intersection(vector<int> &nums1, vector<int> &nums2) {vector<int> res;sort(nums1.begin(), nums1.end());sort(nums2.begin(), nums2.end());int i = 0, j = 0;while (i < nums1.size() && j < nums2.size()) {if (nums1[i] < nums2[j]) ++i;else if (nums1[i] > nums2[j]) ++j;else {res.push_back(nums1[i]);++i;++j;while (nums1[i] == nums1[i - 1]) ++i;while (nums2[j] == nums2[j - 1]) ++j;}}return res;}
};

Discuss

using c++ set

vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {set<int> s(nums1.begin(), nums1.end());vector<int> out;for (int x : nums2)if (s.erase(x))out.push_back(x);return out;
}

急需看完c++ primer 啊!!

Reference

leetcode
c++ set

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leetcode 349. Intersection of Two Arrays 1

Description

My solution

Discuss

Reference

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