523. Continuous Subarray Sum 起止点是K的倍数

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.

map.put剩下的total_sum和index，求到这里来了
之后一直提取if (k != 0) total_sum %= k;

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        //cc
        if (nums == null || nums.length == 0) return false;
        
        //ini: map
        Map<Integer, Integer> map = new HashMap<Integer, Integer>(){{put(0,-1);}};;
        int total_sum = 0;
        
        //for loop: get the same sum
        for (int i = 0; i < nums.length; i++) {
            total_sum += nums[i];
            if (k != 0) total_sum %= k;
            Integer pos = map.get(total_sum);
            if (pos != null) {
                if (i - pos > 1) return true;
            }else {
                map.put(total_sum, i);
            }
        }
        
        return false;
    }
}