523. Continuous Subarray Sum

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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.

题目含义：判断数组中是否有长度至少为2且相加和是k的倍数的子数组

方法一：

 1     public boolean checkSubarraySum(int[] nums, int k) {
 2         for (int i=0;i<nums.length;i++)
 3         {
 4             long sum = nums[i];
 5             for (int j=i+1;j<nums.length;j++)
 6             {
 7                 sum +=nums[j];
 8                 if (sum == k) return true;
 9                 if (k != 0 && sum % k == 0) return true;
10             }
11         }
12         return false;    
13     }

523. Continuous Subarray Sum

标签：style   write   use   amp   i++   ber   you   sum   数组   

原文地址：http://www.cnblogs.com/wzj4858/p/7698440.html