【动态规划】Leetcode编程题解：523. Continuous Subarray Sum Add to List

题目：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 ofk, that is, sums up to n*k where n is also an integer.

样例：

Example 1:

Input: [23, 2, 4, 6, 7],  k=6Output: TrueExplanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.

Example 2:

Input: [23, 2, 6, 4, 7],  k=6Output: TrueExplanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.

这道题目按照题目要求来就行。

解法一：

class Solution {public:    bool checkSubarraySum(vector<int>& nums, int k) {         for(int i = 0; i < nums.size(); i++) {             int num = nums[i];             for(int j = i + 1; j < nums.size(); j++) {                 num += nums[j];                 if(num == k) return true;                 if(k != 0 && num % k == 0) return true;             }         }           return false;    }};

也可以用一个数学上的技巧：若数字a和b分别除以数字c，若得到的余数相同，那么(a-b)必定能够整除c。并且利用hash表保存余数，降低时间复杂度。

解法二：

class Solution {public:    bool checkSubarraySum(vector<int>& nums, int k) {        int n = nums.size(), sum = 0;        unordered_map<int, int> m{{0,-1}};        for (int i = 0; i < n; ++i) {            sum += nums[i];            int t = (k == 0) ? sum : (sum % k);            if (m.count(t)) {                if (i - m[t] > 1) return true;            } else m[t] = i;        }        return false;    }};