[leetcode]-523-Continuous Subarray Sum

题目：

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.

代码：

       public boolean checkSubarraySum(int[] nums, int k) {
        //add (0,-1) to deal with mod/sum is zero (make sure cur-prev > 1)
        Map map = new HashMap(){{put(0,-1);}};;
        int runningSum = 0;
        for (int i=0;i 1) return true;
            }
            else map.put(runningSum, i);
        }
        return false;
     }

 

类似的题目：[leetcode]-525. Contiguous Array

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