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  • 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;
        }
    }
    View Code
     
     
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  • 原文地址:https://www.cnblogs.com/immiao0319/p/13905067.html
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