-
Notifications
You must be signed in to change notification settings - Fork 0
/
Continuous Subarray Sum
97 lines (92 loc) · 2.7 KB
/
Continuous Subarray Sum
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
/*
LeetCode 523:
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.
Solution:
Language: C++
Tag: hashmap
*/
class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
/*
// 1. O(n^2) time + O(1) space
for (int i = 0; i < nums.size(); ++i) {
int sum = nums[i];
for (int j = i + 1; j < nums.size(); ++j) {
sum += nums[j];
if (sum == 0) {
return true;
}
if (k != 0 && sum % k == 0) {
return true;
}
}
}
return false;
*/
/*
// 2. unordered_map -- O(n) time + O(n) space
if (k == 0) {
int i = 0, j = 0;
while (i < nums.size()) {
while (i < nums.size() && nums[i] != 0) {
++i;
}
j = i;
while (j < nums.size() && nums[j] == 0) {
++j;
}
if (j - i >= 2) {
return true;
}
i = j;
}
return false;
}
unordered_map<int, int> hash;
hash.insert({0, 0});
int sum = 0;
for (int i = 1; i <= nums.size(); ++i) {
sum += nums[i - 1];
int tmp = sum % k;
if (hash.find(tmp) != hash.end()) {
if (i - hash[tmp] >= 2) {
return true;
}
} else {
hash.insert({tmp, i});
}
}
return false;
*/
// 3. modify of 2
unordered_map<int, int> hash;
hash.insert({0, -1});
int sum = 0;
for (int i = 0; i < nums.size(); ++i) {
sum += nums[i];
if (k != 0) {
sum %= k;
}
if (hash.find(sum) != hash.end()) {
if (i - hash[sum] >= 2) {
return true;
}
} else {
hash.insert({sum, i});
}
}
return false;
}
};