Comments (2)
tree的遍历类似于dfs的特殊情况。preorder, inorder, postorder相当于对于stack进出栈的不同讨论。
tree是分别处理左右子树, backtrack是处理for循环
from leetcode.
1079. Letter Tile Possibilities
通过Counter来避免生成重复值
class Solution:
def numTilePossibilities(self, tiles: str) -> int:
counter = Counter(tiles)
self.result = 0
def backtracking():
self.result += 1
for i in range(26):
c = chr(ord('A') + i)
if counter[c] == 0:
continue
counter[c] -= 1
backtracking()
counter[c] += 1
backtracking()
return self.result
也可以通过对字符串排序,然后控制index来避免重复
from leetcode.
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from leetcode.