LeetCode-349&350.Intersection of Two Arrays

349 https://leetcode.com/problems/intersection-of-two-arrays/

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.

这是一道有陷阱的简单题

方法1

代码用了两个set存放原始num，为的是去掉重复元素。如果不适用set2，会存在重复元素的问题

vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {unordered_set<int> set1,set2;for (int n : nums1)set1.insert(n);for (int n : nums2)set2.insert(n);vector<int> res;for (int n : set2){if (!set1.insert(n).second)    res.push_back(n);}return res;}

方法2

排序后使用双指针

vector<int> intersection(vector<int>& nums1, vector<int>& nums2)     {        int n1 = nums1.size(), n2 = nums2.size(),i=0,j=0;    vector<int> res;    if (n1 == 0 || n2 == 0)    return res;    sort(nums1.begin(), nums1.end());    sort(nums2.begin(), nums2.end());    while (i<n1&&j<n2)    {    if (nums1[i] < nums2[j])    i++;    else if (nums1[i] > nums2[j])    j++;    else    {    if (find(res.begin(), res.end(), nums1[i]) == res.end())    res.push_back(nums1[i]);    i++;    j++;    }    }    return res;    }

二分查找

bool search(vector<int> nums, int n)    {    int l = 0, r = nums.size() - 1, m;    while (l <= r)    {    m = (l + r) / 2;    if (nums[m] < n)    l = m + 1;    else if (nums[m] > n)    r = m - 1;    else    return true;    }    return false;    }        vector<int> intersection(vector<int>& nums1, vector<int>& nums2)     {        int n1 = nums1.size(), n2 = nums2.size();    vector<int> res;    if (n1 == 0 || n2 == 0)    return res;    sort(nums1.begin(), nums1.end());    for (int n : nums2)    {    if (search(nums1,n)&& find(res.begin(), res.end(), n) == res.end())    res.push_back(n);    }    return res;    }

350 https://leetcode.com/problems/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?

方法1

直接在nums1中查找每一个nums2元素，如果存在就加入到结果列表，并在nums1中删除当前元素

vector<int> intersect(vector<int>& nums1, vector<int>& nums2)     {        int n1 = nums1.size(), n2 = nums2.size();    vector<int> res;    if (n1 == 0 || n2 == 0)    return res;    for (int n : nums2)    {    auto i = find(nums1.begin(), nums1.end(), n);    if ( i != nums1.end())    {    res.push_back(n);    nums1.erase(i);    }    }    return res;    }

方法2

同349一样用双指针，同时不用判断是否存在在结果列表中

vector<int> intersect(vector<int>& nums1, vector<int>& nums2)     {        int n1 = nums1.size(), n2 = nums2.size(), i = 0, j = 0;    vector<int> res;    if (n1 == 0 || n2 == 0)    return res;    sort(nums1.begin(), nums1.end());    sort(nums2.begin(), nums2.end());    while (i < n1&&j < n2)    {    if (nums1[i] < nums2[j])    i++;    else if (nums1[i] > nums2[j])    j++;    else    {    res.push_back(nums1[i]);    i++;    j++;    }    }    return res;    }

方法3

计数法 空间消耗O(m)，如果不判断是否存在，则消耗O(m + n)。说明参考 https://leetcode.com/discuss/103787/table-solution-pointers-solution-with-time-space-complexity 

Based on C++ map mechanism, if a key is not exist, access the key will assign a default value to the key. so if you simply test if map[key] is 0 or not by using "if (map[key] == 0)" without testing if the key is in the map. you will consume extra space....you could avoid allocate extra space either by find or count method. I usually use count, it is more concise. .

vector<int> intersect(vector<int>& nums1, vector<int>& nums2)     {        int n1 = nums1.size(), n2 = nums2.size();    vector<int> res;    if (n1 == 0 || n2 == 0)    return res;    unordered_map<int, int> dict;    for (int n : nums1)     dict[n]++;    for (int n : nums2)    if (dict.find(n) != dict.end() && --dict[n] >= 0)    res.push_back(n);    return res;    }