Given a list ofnon-negativenumbers and a targetintegerk, write a function to check if the array has a continuous subarray of size at least 2 that sums up to a multiple ofk, that is, sums up to n*k where n is also aninteger.

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.

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

O(n^2) 方法，把presum計算出來，然後分k=0/≠0找有沒有合適的解

/** Key point: if we can find any two subarray of prefix sum have same mod value, then their difference MUST be
 * divisible by k. So we can use a map to store mod value of each prefix sum in map, with its index. Then check
 * if map contains the same mod value with size > 2 when we have new mod value in every iteration */
public boolean checkSubarraySum(int[] nums, int k) {
    if (nums.length < 2) {
        return false;
    }

    Map<Integer, Integer> map = new HashMap<>();
    // corner case: if the very first subarray with first two numbers in array could form the result, we need to 
    // put mod value 0 and index -1 to make it as a true case
    map.put(0, -1);
    int curSum = 0;
    for (int i = 0; i < nums.length; i++) {
        curSum += nums[i];

        // corner case: k CANNOT be 0 when we use a number mod k
        if (k != 0) {
            curSum = curSum % k;
        }
        if (map.containsKey(curSum)) {
            if (i - map.get(curSum) > 1) {
                return true;
            }
        }
        else {
            map.put(curSum, i);
        }
    }
    return false;
}

把presum 對 k 的mod和它的first index存到map裡，為了應對k==0，要map.put(0, -1)，然後在迴圈中持續檢查即可