-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy path1221-SplitAStringInBalancedStrings.go
More file actions
63 lines (53 loc) · 2.01 KB
/
1221-SplitAStringInBalancedStrings.go
File metadata and controls
63 lines (53 loc) · 2.01 KB
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
package main
// 1221. Split a String in Balanced Strings
// Balanced strings are those that have an equal quantity of 'L' and 'R' characters.
// Given a balanced string s, split it into some number of substrings such that:
// Each substring is balanced.
// Return the maximum number of balanced strings you can obtain.
// Example 1:
// Input: s = "RLRRLLRLRL"
// Output: 4
// Explanation: s can be split into "RL", "RRLL", "RL", "RL", each substring contains same number of 'L' and 'R'.
// Example 2:
// Input: s = "RLRRRLLRLL"
// Output: 2
// Explanation: s can be split into "RL", "RRRLLRLL", each substring contains same number of 'L' and 'R'.
// Note that s cannot be split into "RL", "RR", "RL", "LR", "LL", because the 2nd and 5th substrings are not balanced.
// Example 3:
// Input: s = "LLLLRRRR"
// Output: 1
// Explanation: s can be split into "LLLLRRRR".
// Constraints:
// 2 <= s.length <= 1000
// s[i] is either 'L' or 'R'.
// s is a balanced string.
import "fmt"
func balancedStringSplit(s string) int {
balance, res := 0, 0
for i, ch := range s {
if ch == 'L' { balance++ }
if ch == 'R' { balance-- }
if i != 0 && balance == 0 { // 能够抵消的算一次
res++
}
}
return res
}
func main() {
// Example 1:
// Input: s = "RLRRLLRLRL"
// Output: 4
// Explanation: s can be split into "RL", "RRLL", "RL", "RL", each substring contains same number of 'L' and 'R'.
fmt.Println(balancedStringSplit("RLRRLLRLRL")) // 4
// Example 2:
// Input: s = "RLRRRLLRLL"
// Output: 2
// Explanation: s can be split into "RL", "RRRLLRLL", each substring contains same number of 'L' and 'R'.
// Note that s cannot be split into "RL", "RR", "RL", "LR", "LL", because the 2nd and 5th substrings are not balanced.
fmt.Println(balancedStringSplit("RLRRRLLRLL")) // 2
// Example 3:
// Input: s = "LLLLRRRR"
// Output: 1
// Explanation: s can be split into "LLLLRRRR".
fmt.Println(balancedStringSplit("LLLLRRRR")) // 1
}