题目
You are given two strings order and s. All the characters of order are unique and were sorted in some custom order previously.
Permute the characters of s so that they match the order that order was sorted. More specifically, if a character x occurs before a character y in order, then x should occur before y in the permuted string.
Return any permutation of s that satisfies this property.
解题思路
方法一，用字典给每个字母一个数作为大小比较，没出现在order里的默认为-1，自定义排序比较函数
方法二，用counter求出每个字母出现的次数，遍历order，如果遇到出现的字母，就产生对此个数的字母，把所有没出现在order里的放到最后
代码
class Solution:
def customSortString(self, order: str, str: str) -> str:
ct = Counter(str)
res = []
for c in order:
res.append(c * ct[c])
ct[c] = 0
for c, v in ct.items():
res.append(c * v)
return ''.join(res)
class Solution:
def customSortString(self, order: str, str: str) -> str:
d = {c: i for i, c in enumerate(order)}
return ''.join(sorted(str, key=lambda x: d.get(x, -1)))
