**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。
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)
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)