523. Continuous Subarray Sum是否有连续和是某数的几倍

［抄题］：

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.

 [暴力解法]：

时间分析：

空间分析：

 [优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

［思维问题］：

(a+(n*x))%x is same as (a%x)

For e.g. in case of the array [23,2,6,4,7] the running sum is [23,25,31,35,42] and the remainders are [5,1,1,5,0]. We got remainder 5 at index 0 and at index 3. That means, in between these two indexes we must have added a number which is multiple of the k. Hope this clarifies your doubt :)

［一句话思路］：

余数重复，必然增加了几倍

［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

1. pos是之前的index,应该用i - pos是否>1来检测

［二刷］：

［三刷］：

［四刷］：

［五刷］：

  [五分钟肉眼debug的结果]：

［总结］：

［复杂度］：Time complexity: O(n) Space complexity: O(n)

［英文数据结构或算法，为什么不用别的数据结构或算法］：

［算法思想：递归/分治/贪心］：

［关键模板化代码］：

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

 [代码风格] ：

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;
    }
}