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:
- The length of the array won‘t exceed 10,000.
- 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 }
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