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=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:
- 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.
preSum一下,然后做差判断是否符合条件
代码:
public boolean checkSubarraySum(int[] nums, int k) { if(nums == null || nums.length < 2) return false; int[] preSum = new int[nums.length+1]; int sum = 0; for(int i=1;i<=nums.length;i++) { sum += nums[i-1]; preSum[i] = sum; } for(int i=0;i<nums.length;i++) { for(int j=i+2;j<=nums.length;j++) { if(k == 0 && ((preSum[j] == preSum[i]))) return true; if(k != 0 && (((preSum[j] - preSum[i]) % k) == 0)) return true; } } return false; }
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