1. Description
2. Solution
**解析：**两种思路，一种思路是对所有的.，判断是否替换，如果需要替换，根据可能的情况分析替换成L还是R，通过左右双指针实现。一种思路是对所有的L和R，替换其附近需要替换的.，首先，对于L左边没有R的情况，替换.为L，对于R右边不存在L的情况，替换.为R；对于R右边存在L的情况，R和L正中间的.保持不变，左半部分变为R，右边部分变为L，循环从L的下一位重新开始。
- Version 1
class Solution:
def pushDominoes(self, dominoes: str) -> str:
n = len(dominoes)
state = list(dominoes)
for i in range(n):
if state[i] == '.':
left = i - 1
right = i 1
while left > -1 and dominoes[left] == '.':
left -= 1
while right < n and dominoes[right] == '.':
right = 1
if left != -1 and right != n:
if dominoes[left] == 'R' and dominoes[right] == 'L':
if i - left > right - i:
state[i] = 'L'
elif i - left < right - i:
state[i] = 'R'
elif dominoes[left] == 'R' and dominoes[right] == 'R':
state[i] = 'R'
elif dominoes[left] == 'L' and dominoes[right] == 'L':
state[i] = 'L'
elif right != n and dominoes[right] == 'L':
state[i] = 'L'
elif left != -1 and dominoes[left] == 'R':
state[i] = 'R'
return ''.join(state)- Version 2
class Solution:
def pushDominoes(self, dominoes: str) -> str:
n = len(dominoes)
state = list(dominoes)
i = 0
while i < n:
if dominoes[i] == 'L':
left = i - 1
while left > -1 and dominoes[left] == '.':
state[left] = 'L'
left -= 1
elif dominoes[i] == 'R':
right = i 1
while right < n and dominoes[right] == '.':
state[right] = 'R'
right = 1
if right != n and dominoes[right] == 'L':
for k in range(right - (right - i - 1) // 2, right):
state[k] = 'L'
if (right - i - 1) % 2 == 1:
state[i (right - i - 1) // 2 1] = '.'
i = right
i = 1
return ''.join(state)Reference
- https://leetcode.com/problems/push-dominoes/
