350. Intersection of Two Arrays II

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?

解答： 题目很容易理解，就是在两个数组中找相同的元素。这里有很多的解法，我都会粘贴上来。

值得高兴的是，我的方法用时8ms, 击败86%，哈哈~ 不过貌似简单的方法也能达到如此。

1.用二分查找，这里受  278. First Bad Version 的启发，将二分查找修改使得适用于能返回重复元素找到第一个位置。

同时符合follow up中第三条，对每个nums2中的元素遍历获取，然后从nums1中去查找有无相同元素，同时用一个标记数组标记num1，使得若找到，则判断该位置上是否已经被获取过，若被获取过，则到下一个位置判断有没有获取过，直到下一个位置没被获取过（则加入结果集合中）或者与想要的元素值不同（跳过）。

class Solution {public:    int searchfirst(int num, vector<int>& vec){    int begin = 0;    int end = vec.size()-1;    int middle;    while(begin < end - 1){        middle = begin+(end-begin)/2;        if(vec[middle] >= num){            end = middle;        }else            begin = middle;    }    if(vec[begin] == num) return begin;    else return end;}vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {    int len1 = nums1.size();    sort(nums1.begin(), nums1.end());    vector<bool>v1(len1, false);    vector<int>res;    int len2 = nums2.size();    if(len1 == 0 || len2 == 0)    return res;    int i = 0;    while(i < len2){        int temp = nums2[i];        int index = searchfirst(temp,nums1);        while(nums1[index] == temp){            if(v1[index] == false){                res.push_back(temp);                v1[index]=true;                break;            }else                index++;        }        ++i;    }    return res;}};

下面粘贴其他的方法

2.对两个数组都进行排序，然后进行比较（这个基于两个数组都能一次性download下来）

Sort and two pointers Solution: Time: O(max(m, n) log(max(m, n))) Space: O(m + n)

vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {    vector<int> result;    result.reserve(min(nums1.size(), nums2.size()));    sort(nums1.begin(), nums1.end());    sort(nums2.begin(), nums2.end());    int i1 = 0, i2 = 0, len1 = nums1.size(), len2 = nums2.size();    while(i1 < len1 && i2 < len2)    {        if(nums1[i1] == nums2[i2])        {            result.push_back(nums1[i1]);            ++i1; ++i2;        }        else if(nums1[i1] < nums2[i2])            ++i1;        else            ++i2;    }    return result;}

3.使用hashmap

(1)m: nums1.size n: nums2.size

Hash table solution: Time: O(m + n) Space: O(m + n)

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

(2)Hash table solution2: Time: O(m + n) Space: O(m)

class Solution {public:    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {        unordered_map<int, int> dict;        vector<int> res;        for(int i = 0; i < (int)nums1.size(); i++) dict[nums1[i]]++;        for(int i = 0; i < (int)nums2.size(); i++)            if(dict.find(nums2[i]) != dict.end() && --dict[nums2[i]] >= 0) res.push_back(nums2[i]);        return res;    }};