1221. Split a String in Balanced Strings*
https://leetcode.com/problems/split-a-string-in-balanced-strings/
题目描述
Balanced strings are those who have equal quantity of 'L' and 'R' characters.
Given a balanced string s split it in the maximum amount of balanced strings.
Return the maximum amount of splitted balanced strings.
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 = "RLLLLRRRLR"
Output: 3
Explanation: s can be split into "RL", "LLLRRR", "LR", each substring contains same number of 'L' and 'R'.
Example 3:
Input: s = "LLLLRRRR"
Output: 1
Explanation: s can be split into "LLLLRRRR".
Example 4:
Input: s = "RLRRRLLRLL"
Output: 2
Explanation: s can be split into "RL", "RRRLLRLL", since each substring contains an equal number of 'L' and 'R'
Constraints:
-
1 <= s.length <= 1000 -
s[i] = 'L' or 'R'
C++ 实现 1
使用 Hash 表, 遍历字符串, 记录 L 和 R 的数量, 如果数量相等, 那么进行 split, 按照这种方式进行 split 可以获得 maximum amount.
class Solution {
public:
int balancedStringSplit(string s) {
int count = 0;
unordered_map<char, int> record;
for (int i = 0; i < s.size(); ++ i) {
record[s[i]] ++;
if (record['R'] == record['L']) {
count ++;
}
}
return count;
}
};
C++ 实现 2
这里使用 candicate 统计 L 或 R 的个数 (个数用 indicator 表示). 只有在 indicator == 0 时, 表明 L 和 R 的个数相等.
class Solution {
public:
int balancedStringSplit(string s) {
int count = 0, indicator = 0;
char candicate = '\0';
for (int i = 0; i < s.size(); ++ i) {
if (candicate == '\0') {
candicate = s[i];
}
if (s[i] == candicate) indicator += 1;
else indicator -= 1;
if (indicator == 0) {
count += 1;
candicate = '\0';
}
}
return count;
}
};
