[leetcode]-523-Continuous Subarray Sum

題目：

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 a 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.

思路：

We iterate through the input array exactly once, keeping track of the running sum mod k of the elements in the process. If we find that a running sum value at index j has been previously seen before in some earlier index i in the array, then we know that the sub-array (i,j)contains a desired sum.

代碼：

       public boolean checkSubarraySum(int[] nums, int k) {
        //add (0,-1) to deal with mod/sum is zero (make sure cur-prev > 1)
        Map<Integer, Integer> map = new HashMap<Integer, Integer>(){{put(0,-1);}};;
        int runningSum = 0;
        for (int i=0;i<nums.length;i++) {
            runningSum += nums[i];
            // take care for k == 0
            if (k != 0) runningSum %= k;
            // System.out.println("running sum: " + runningSum);
            Integer prev = map.get(runningSum);
            // System.out.println("prev: " + prev);
            if (prev != null) {
                if (i - prev > 1) return true;
            }
            else map.put(runningSum, i);
        }
        return false;
     }

 

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https://blog.csdn.net/ljh0302/article/details/104161139