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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.

題目含義:判斷數組中是否有長度至少為2且相加和是k的倍數的子數組

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