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leetcode刷題,總結,記錄,備忘 350

leetcode350Intersection 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?

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Intersection of Two Arrays的第一題比較相似,就是允許重複的,稍微改下程式碼就行了。
class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        
        vector<int> result;
        
        int size1 = nums1.size();
        int size2 = nums2.size();
        
        int i = 0;
        int j = 0;
        
        while (i != size1 && j != size2)
        {
            if (nums1[i] < nums2[j])
            {
                i++;
            }
            else if (nums1[i] > nums2[j])
            {
                j++;
            }
            else if (nums1[i] == nums2[j])
            {
                result.push_back(nums1[i]);
                i++;
                j++;
            }
            else
            {
                i++;
                j++;
            }
        }
        
        return result;
    }
};
我還看到一個比較巧妙的解決方法,用map記錄第一個陣列中每個數字出現的次數,然後在第二個陣列中通過數字做鍵,將數字出現次數做自減,根據結果是否小於0來判斷是否出現在2個數組中。
class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> result;
        
        map<int, int> mi;
        
        for (int i = 0; i < nums1.size(); ++i)
        {
            mi[nums1[i]]++;
        }
        
        for (int i = 0; i < nums2.size(); ++i)
        {
            if (--mi[nums2[i]] >= 0)
            {
                result.push_back(nums2[i]);
            }
        }
        
        return result;
    }
};