leetcode 523. Continuous Subarray Sum 子数组和 + 动态规划DP

水深无声 2022-06-03 03:18 179阅读 0赞

Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n\*k where n is also an integer.

Example 1:  
Input: \[23, 2, 4, 6, 7\], k=6  
Output: True  
Explanation: Because \[2, 4\] is a continuous subarray of size 2 and sums up to 6.  
Example 2:  
Input: \[23, 2, 6, 4, 7\], k=6  
Output: True  
Explanation: Because \[23, 2, 6, 4, 7\] is an continuous subarray of size 5 and sums up to 42.  
Note:  
The length of the array won’t exceed 10,000.  
You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

板梯题意很简单，就在一个暴力遍历即可

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    
    using namespace std;
    
    class Solution 
    {
    public:
        bool checkSubarraySum(vector<int>& nums, int k) 
        {
            vector<int> dp(nums.size()+1, 0);
            for (int i = 1; i <= nums.size(); i++)
            {
                dp[i] = dp[i-1] + nums[i-1];
            }
    
            for (int i = 0; i < nums.size(); i++)
            {
                for (int j = i + 2; j <= nums.size(); j++)
                {
                    if (k == 0)
                    {
                        if (dp[j] - dp[i] == 0)
                            return true;
                    }
                    else if ((dp[j] - dp[i]) % k == 0)
                        return true;
                }
            }
            return false;
        }
    };