Leetcode 523.Continuous Subarray Sum

Leetcode 523.Continuous Subarray Sum

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 the 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.

Note:

  1. The length of the array won’t exceed 10,000.
  2. You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

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.

乍一看没看懂上面的英语,其实它就是这个意思:
对一个数组【a1,a2,a3,a4,a5…am…an…】,用A表示数组每一个元素的累加和
Am = a1 +a2 +a3 +…am
An = a1 +a2 +a3 +…an
如果 Am % k == An % k ,则(An-Am) % k= 0。
说明Am到An之间的数字的和,除以k的余数也为0。
那么,这些数字就是我们要找的一个连续的子串。

java 代码如下:

class Solution {

    public boolean checkSubarraySum(int[] nums, int k) {        
        if(nums.length == 1) return false;
        if(k == 0) {
            for(int i = 0; i < nums.length; i++) {
                if(nums[i] != 0) return false;
            }
            return true;
        }

        HashMap<Integer, Integer> map = new HashMap(); // 余数 : 索引下标
        map.put(0, -1);        
        int sum = 0;        
        for(int i = 0; i < nums.length; i++) {
            sum += nums[i];
            sum %= k;   
            if(map.containsKey(sum)) {
                if(i - map.get(sum) > 1) return true;
            }else {
                map.put(sum, i);
            }
        }
        return false;
    }
}

注:学渣心里苦,不要学楼主,平时不努力,考试二百五,哭 ~

这里写图片描述

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