题目：原题链接（中等）
标签：动态规划、双指针
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( N ) O(N) O(N) | O ( N ) O(N) O(N) | 192ms (37.89%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
class Solution:
def pushDominoes(self, dominoes: str) -> str:
dominoes = list(dominoes)
size = len(dominoes)
lefts = {i for i in range(size) if dominoes[i] == "L"}
rights = {i for i in range(size) if dominoes[i] == "R"}
while lefts or rights:
new_lefts = set()
for left in lefts:
if left == 0:
continue
if left > 0 and (dominoes[left - 1] == "R" or dominoes[left - 1] == "L"):
continue
if left > 1 and dominoes[left - 2] == "R":
continue
else:
dominoes[left - 1] = "L"
new_lefts.add(left - 1)
new_rights = set()
for right in rights:
if right == size - 1:
continue
if right < size - 1 and (dominoes[right + 1] == "L" or dominoes[right + 1] == "R"):
continue
if right < size - 2 and dominoes[right + 2] == "L":
if right + 2 in new_lefts:
dominoes[right + 1] = "R"
else:
dominoes[right + 1] = "R"
new_rights.add(right + 1)
lefts, rights = new_lefts, new_rights
return "".join(dominoes)
