[leetcode-523-Continuous Subarray Sum]

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:

1. The length of the array won't exceed 10,000.
2. You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

思路：

首先想到用动态规划，使用dp[i][j]记录nums中i到j的和，然后随时判断是否满足余数为0.

注意处理k==0时候的情况。

但是直接用二维数组记录的话，提示内存不足，耗费空间。

更新公式为 dp[i][j] = dp[i][j-1]+nums[j] ，可以看出dp[i][j]只与上一个dp[i][j-1]有关，

于是用两个变量替代即可。得到如下代码，时间复杂度为O(n²).

bool checkSubarraySum(vector<int>& nums, int k)
     {
         int len = nums.size();
        // vector<vector<int>> dp(len, vector<int>(len, 0));
         long long cur = , pre = ;
         for (int i = ; i < len;i++)
         {
             for (int j = i; j < len;j++)
             {
                 if (j == i)cur = nums[i];
                 else
                 {
                     cur = pre + nums[j];
                     if (k !=  && cur % k == )return true;
                     else if (k ==  && cur == ) return true;
                 }
                 pre = cur;
             }
         }
         return false;
     }

后来参考网上大牛的代码，学习到了他们的解法。

比如他们用一个set去存储i之前元素和的余数，如果往后遍历到j元素和的余数之前出现过，说明i到j的和为k的整数倍。

这样时间复杂度为O(n).

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        int n = nums.size(), sum = , pre = ;
        unordered_set<int> modk;
        for (int i = ; i < n; ++i) {
            sum += nums[i];
            int mod = k ==  ? sum : sum % k;
            if (modk.count(mod)) return true;
            modk.insert(pre);
            pre = mod;
        }
        return false;
    }
};

参考：

https://discuss.leetcode.com/topic/80892/concise-c-solution-use-set-instead-of-map