Description

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.

Solution

Pre-sum + Two-pointer, time O(n ^ 2), space O(n)

注意要對(duì)k == 0的情況單獨(dú)處理！x % 0是非法的。另外k也可能為負(fù)數(shù)哦，別忘記了。

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        if (nums == null || nums.length < 2) {
            return false;
        }
        
        int n = nums.length;
        int[] sum = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            sum[i + 1] = sum[i] + nums[i];
        }
        
        for (int i = 0; i <= n - 2; ++i) {
            for (int j = i + 2; j <= n; ++j) {
                if ((k == 0 && sum[i] == sum[j]) || (k != 0 && (sum[j] - sum[i]) % k == 0)) {
                    return true;
                }
            }
        }
        
        return false;
    }
}

Pre-sum + HashMap, time O(n), space O(n)

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        if (nums == null || nums.length < 2) {
            return false;
        }
        
        Map<Integer, Integer> sum2Idx = new HashMap<>();
        sum2Idx.put(0, -1);
        int sum = 0;
        
        for (int i = 0; i < nums.length; ++i) {
            sum += nums[i];
            if (k != 0) {
                sum %= k;   // tricky
            }
            
            if (sum2Idx.containsKey(sum)) {
                if (i - sum2Idx.get(sum) > 1) {
                    return true;
                }
            } else {
                sum2Idx.put(sum, i);
            }
        }        
        
        return false;
    }
}