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=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.

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.

似乎没怎么费工夫就完成了，要点应该是时间吧。。。

也就是说这道题是要遍历所有的连续子串的，用L表示The length of the array，那么一共有(L-1)*L/2个连续子串，计算每一个子串的和S，再判断是否有S能被k整除，就是目的了。

关键就在于计算的次数，利用空间换区时间，每次都将上一步的计算结果存储起来，即如果计算连续子串A[i,j]的和，可以用A[i,j] = A[i,j-1] + A[j]的式子得出结果，从A[0,1]开始计算A[0,1] = A[0] + A[1]， 将结果保存，A[0,2] = A[0,1] + A[2]，这样每一步求和计算所用到的加法次数为1。

若不使用额外空间保存则最坏情况的一次求和加法次数为L-1。

最后再将特殊情况加入，当k=0时，不可以使用求余%，那么只要判断数组内是否存在连续两个数为0就可以了，存在则return true，否则return false

代码：

    bool checkSubarraySum(vector<int>& nums, int k) {        int s = nums.size();        if(k == 0){            for(int i = 0; i < s-1; ++i) {                if(nums[i] == 0 && nums[i+1] == 0){                    return true;                }            }            return false;        }        int **a;        a = new int*[s];        for(int i = 0;i < s-1; ++i){            a[i] = new int[s];            for(int j = i+1; j < s; ++j){                if(j - i == 1){                    a[i][j] = (nums[i] + nums[j]) % k;                }                else{                    if(i == 0){                        a[i][j] = (a[i][j-1] + nums[j]) % k;                        }                    else{                        a[i][j] = (a[i-1][j] + k - nums[i]) % k;                    }                }                if(a[i][j] == 0) {                    return true;                }            }                        }        return false;    }