1221. segmentation balance string

1221. segmentation balance string

description:

In a "balanced string", the number of 'L' and 'R' character is the same.

Given a balanced string s, you divide it into as many balance strings.

Returns the maximum number of equilibrium can be obtained by a division of the string.

 

Example 1:

Input: s = "RLRRLLRLRL"
Output: 4
explains: S may be divided into "RL", "RRLL", "RL", "RL", each sub-string contains the same number of 'L' and 'R'.
Example 2:

Input: s = "RLLLLRRRLR"
Output: 3
explains: S may be divided into "RL", "LLLRRR", "LR", each sub-string contains the same number of 'L' and 'R'.
Example 3:

Input: s = "LLLLRRRR"
Output: 1
explains: S can remain as "LLLLRRRR".

Example 4:

Input: s = "RRLRRLRLLLRL"
Output: 2

prompt:

1 <= s.length <= 1000
s[i] = 'L' 或 'R'

 

1 Solution:     0 MS     8.4 MB
 2  class Solution {
 . 3  public :
 . 4      int balancedStringSplit ( String S) {
 . 5          / * ideas:
 . 6              1: Traversal string, index = 0 starts, before the number of records R or L sumx, sumy ;
 . 7             2: current statistics L and R, once equal sum ++, sumx = 0, sumy = 0; can continue to the next counted
 . 8          * / 
. 9          int SUM = 0 ;
 10          int SUMX = 0 , Sumy = 0 ;
 . 11          for ( int = I 0;i<s.size();i++){
12             if(s[i]=='L') sumx++;
13             else sumy++;
14             if(sumx==sumy){
15                 sum++;
16                 sumx=0,sumy=0;
17             }
18         }
19         return sum;
20     }
21 };