如果是'I'就加入当前最小的,是'D'就加入当前最大的.
from collections import Counter, deque class Solution: def diStringMatch(self, S: str) -> List[int]: ans = [] A = deque(range(len(S) + 1)) for s in S: if s=='I': ans.append(A.popleft()) else: ans.append(A.pop()) ans.append(A.pop()) return ans