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Leetcode 523: Continuous Subarray Sum

时间:2018-01-28 12:49:20      阅读:173      评论:0      收藏:0      [点我收藏+]

标签:[]   time   ati   multi   because   sig   contain   may   cti   

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.
 1 public class Solution {
 2     // idea: https://discuss.leetcode.com/topic/80793/java-o-n-time-o-k-space
 3     public bool CheckSubarraySum(int[] nums, int k) {
 4         var dict = new Dictionary<int, int>();
 5         dict[0] = -1;
 6         var sum = 0;
 7         
 8         for (int i = 0; i < nums.Length; i++)
 9         {
10             sum += nums[i];
11             
12             var prev = k == 0 ? sum : sum % k;
13             
14             if (dict.ContainsKey(prev))
15             {            
16                 if (i - dict[prev] > 1) return true;
17             }
18             else
19             {
20                 dict[prev] = i;    
21             }         
22         }
23         
24         return false;
25     }
26 }

 

Leetcode 523: Continuous Subarray Sum

标签:[]   time   ati   multi   because   sig   contain   may   cti   

原文地址:https://www.cnblogs.com/liangmou/p/8370740.html

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