贪心，每次都尽量取大的，除非连续取的次数超出限制，此时取一个下一个字符

class Solution:
    def repeatLimitedString(self, s: str, repeatLimit: int) -> str:
        N = 26
        count = [0] * N
        for c in s:
            count[ord(c) - ord('a')] += 1
        ret = []
        i, j, m = N - 1, N - 2, 0
        while i >= 0 and j >= 0:
            if count[i] == 0: # 当前字符已经填完，填入后面的字符，重置 m
                m, i = 0, i - 1
            elif m < repeatLimit: # 当前字符未超过限制
                count[i] -= 1
                ret.append(chr(ord('a') + i))
                m += 1
            elif j >= i or count[j] == 0: # 当前字符已经超过限制，查找可填入的其他字符
                j -= 1
            else: # 当前字符已经超过限制，填入其他字符，并且重置 m
                count[j] -= 1
                ret.append(chr(ord('a') + j))
                m = 0
        return ''.join(ret)