[leetcode]523. Continuous Subarray Sum连续子数组和(为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 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.

 

题目

给定数组和一个数K，求是否存在子数组和为K的倍数。

 

思路

 

代码

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        HashMap<Integer, Integer> map = new HashMap<>();
        //为何 map.put(0, -1) 呢？ 如果在第2位找到了mod == 0的数，那就 1 -（-1）>1，return true。
        map.put(0, -1);
        int sum = 0;
        for (int i = 0; i < nums.length; i++) {
            // running sum 
            sum += nums[i];
            if (k != 0) {
                sum %= k;
            }
            // find sum % k is in the HashMap
            if (map.containsKey(sum)) {
                // subarray length at least two 
                if (i - map.get(sum) > 1) {
                    return true;
                }
            }
            else {
                // key: runnng sum -- value: index
                map.put(sum, i);
            }
        }
        return false;
    }
}