leetcode 523. Continuous Subarray Sum

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本文介绍了一种算法，用于判断给定整数数组中是否存在至少包含两个元素的连续子数组，其和为给定整数k的倍数。通过使用哈希映射记录累积和及其对应的索引来实现这一目标。

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

利用一个HashMap存放加到之前结果模上k的结果和位置
如果该结果再次出现，说明之前出现位置之后到当前位置的序列是满足要求的序列。

public class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        if(nums==null||nums.length==0) return false;
        Map<Integer, Integer> map = new HashMap<Integer, Integer>(){{put(0,-1);}};;
        int runningSum = 0;
        for (int i=0;i<nums.length;i++) {
            runningSum += nums[i];
            if (k != 0) runningSum %= k; 
            Integer prev = map.get(runningSum);
            if (prev != null) {
                if (i - prev > 1) return true;
            } else map.put(runningSum, i);
        }
        return false;
    }
}