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

Example 1:

Input: nums1 = [1,2,2,1], nums2 = [2,2]Output: [2,2]

Example 2:

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]Output: [4,9]

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?

 

 

这道题是之前那道 的拓展，不同之处在于这道题允许我们返回重复的数字，而且是尽可能多的返回，之前那道题是说有重复的数字只返回一个就行。那么这道题我们用哈希表来建立nums1中字符和其出现个数之间的映射, 然后遍历nums2数组，如果当前字符在哈希表中的个数大于0，则将此字符加入结果res中，然后哈希表的对应值自减1，参见代码如下：

 

解法一：

class Solution {public:    vector
     
       intersect(vector
      
       & nums1, vector
       
        & nums2) {        unordered_map
        
          m;        vector
         
           res;        for (auto a : nums1) ++m[a];        for (auto a : nums2) {            if (m[a]-- > 0) res.push_back(a);        }        return res;    }};
         
        
       
      
     

 

再来看一种方法，这种方法先给两个数组排序，然后用两个指针分别指向两个数组的起始位置，如果两个指针指的数字相等，则存入结果中，两个指针均自增1，如果第一个指针指的数字大，则第二个指针自增1，反之亦然，参见代码如下：

 

解法二：

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

 

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