523. Continuous Subarray Sum

description:

 523

Continuous Subarray Sum   

23.2%Medium

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.

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.

thought:

这道题没想出什么创新的方法，所以就不想po自己的代码上去了，看了一下别人的solution，发现可以先把和求出来，然后再相减的方法，进行连续子串和的计算，然后取模，感觉是一种比较巧妙地方法了。

Better ways:

by tankztc

class Solution {public:    bool checkSubarraySum(vector<int>& nums, int k) {        if (nums == null || nums.length == 0)   return false;                int[] preSum = new int[nums.length+1];                for (int i = 1; i <= nums.length; i++) {            preSum[i] = preSum[i-1] + nums[i-1];        }                for (int i = 0; i < nums.length; i++) {            for (int j = i+2; j <= nums.length; j++) {                if (k == 0) {                    if (preSum[j] - preSum[i] == 0) {                        return true;                    }                } else if ((preSum[j] - preSum[i]) % k == 0) {                    return true;                }            }        }        return false;    }};