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题目：（简单）

标签：字符串

解法一（循环处理）：

def countAndSay(self, n: int) -> str:    def count(s):        new = ""        last = None        num = 0        for c in s:            if c != last:                if num > 0:                    new += str(num) + last                last = c                num = 1            else:                num += 1        if num > 0:            new += str(num) + last        return new    s = "1"    for i in range(1, n):        s = count(s)    return s

解法二（递归实现）：

def countAndSay(self, n: int) -> str:    def count(s):        new = ""        last = None        num = 0        for c in s:            if c != last:                if num > 0:                    new += str(num) + last                last = c                num = 1            else:                num += 1        if num > 0:            new += str(num) + last        return new    def get_num(n):        if n == 1:            return "1"        else:            return count(get_num(n - 1))    return get_num(n)

解法三（枚举法）：

class Solution:    def countAndSay(self, n: int) -> str:        count_dict = {
   1: '1', 2: '11', 3: '21', 4: '1211', 5: '111221', 6: '312211', 7: '13112221', 8: '1113213211', 9: '31131211131221', 10: '13211311123113112211',......}        return count_dict[n]

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