523. Continuous Subarray Sum 

 1 //New presum
 2 class Solution {
 3     public boolean checkSubarraySum(int[] nums, int k) {
 4         if(k == 0){
 5             for(int i = 1; i < nums.length; i++){
 6                 if(nums[i] == 0 && nums[i-1] == 0) return true;
 7             }
 8             return false;
 9         }
10         int[] presum = new int[nums.length + 1];
11         for(int i = 0; i < nums.length; i++){
12             presum[i+1] = presum[i] + nums[i];
13         }
14         for(int i = 0; i <= nums.length-2; i++){
15             for(int j = i+2; j <= nums.length; j++){
16                 if((presum[j] - presum[i]) % k == 0){
17                     return true;
18                 }
19             }
20         }
21         return false;
22         
23     }
24 }
25 
26 //Old
27 class Solution {
28     public boolean checkSubarraySum(int[] nums, int k) {
29         
30         List<Integer> list = new ArrayList<>();
31         for(int i = 0; i < nums.length; i++){
32             for(int j = 0; j < list.size(); j++){
33                 // System.out.println(list.get(j) + nums[i]);
34                 if((k == 0 && list.get(j) + nums[i] == 0) || (k != 0 && (list.get(j) + nums[i]) % k == 0)){
35                     return true;
36                 }else{
37                     list.set(j, list.get(j) + nums[i]);
38                 }
39             }
40             list.add(nums[i]);
41         }
42         return false;
43         
44     }
45 }

 

posted @ 2018-10-23 05:27  jasoncool1  阅读(135)  评论(0)    收藏  举报