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:

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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:

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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.

解法1：O(N^2)

Subarray Sum的题目考虑用prefix sum解决，先计算每一个prefix sum，然后再遍历每一个subarray来寻找满足条件的最大答案。
C++

1

Java

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public class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        if (nums == null || nums.length == 0) {
            return false;
        }
        // prefix sum
        int[] dp = new int[nums.length + 1];
        for (int i = 1; i <= nums.length; i++) {
            dp[i] = dp[i - 1] + nums[i - 1];
        }
        // check if there is a sum - prefix that is n * k
        for (int i = 1; i <= nums.length; i++) {
            for (int j = 0; j < i - 1; j++) {
                int temp = dp[i] - dp[j];
                if (k == 0 && temp == 0) {
                    return true;
                } else if (k != 0 && temp % k == 0) {
                    return true;
                }
            }
        }
        return false;
    }
}