2038. Remove Colored Pieces if Both Neighbors are the Same Color
两个人负责A跟B
轮流把字串中 各自负责的字母连续三个变成连续两个
谁不能操作谁就输 永远是A先手
思路：
这题不是博弈题
只要轮流操作计次
判断最后A次数是否大于B次数就好
Code:
impl Solution {
pub fn winner_of_game(colors: String) -> bool {
let mut a_count = 0;
let mut b_count = 0;
let bytes = colors.as_bytes();
for index in 2..colors.len() {
if bytes[index] == bytes[index - 1] &&
bytes[index] == bytes[index - 2] {
if bytes[index] == b'A' {
a_count += 1;
} else {
b_count += 1;
}
}
}
(a_count > b_count)
}
}