以s的长度+1为最大数，以0为最小数。遍历s，遇到I，res贴最小数字，最小数+1，遇到D，贴最大数，最大数-1.

class Solution:
    def diStringMatch(self, s: str) -> List[int]:
        N = len(s)
        res = []
        small = 0
        big = N 
        for idx, char in enumerate(s):
            if char == 'I':
                res.append(small)
                small += 1
            elif char == 'D':
                res.append(big)
                big -= 1
        if s[-1] == 'D':
            res.append(big)
        elif s[-1] =='I':
            res.append(small)
        return res