[Leetcode][Python]19: Remove Nth Node From End of List

最新推荐文章于 2025-12-16 10:44:31 发布

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原文链接：http://www.cnblogs.com/Dabay/p/4276253.html

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#python

本文解析了LeetCode上的经典题目Count and Say（编号38），详细介绍了如何通过递归的方式生成著名的Count and Say数列，并提供了Python实现代码。Count and Say数列从数字1开始，后续每一项都是前一项的读法描述。

# -*- coding: utf8 -*-
'''
__author__ = 'dabay.wang@gmail.com'

38: Count and Say
https://oj.leetcode.com/problems/count-and-say/

The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 111221, ...

1 is read off as "one 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.
Given an integer n, generate the nth sequence.
Note: The sequence of integers will be represented as a string.

===Comments by Dabay===
题意半天没搞懂。原来是给的数字n是几，就返回第几个字符串。例如，如果n是5，就返回“111221”这个字符串。
第一个铁定是1，然后用say的方式来往后生成下一个字符串。

say的时候：
比较下一个数字是否一样，
    如果一样，计数器加一
    如果不一样，say
'''

class Solution:
    # @return a string
    def countAndSay(self, n):
        current_result = "1"
        start = 1
        while start < n:
            previous_result = current_result
            current_result = ""
            counting_number = None
            counter = 0
            for num in previous_result:
                if counting_number is None:
                    counting_number = num
                    counter = 1
                elif counting_number == num:
                    counter += 1
                else:
                    current_result += "%s%s" % (counter, counting_number)
                    counting_number = num
                    counter = 1
            else:
                current_result += "%s%s" % (counter, counting_number)
            start += 1
        return current_result


def main():
    sol = Solution()
    print sol.countAndSay(5)


if __name__ == "__main__":
    import time
    start = time.clock()
    main()
    print "%s sec" % (time.clock() - start)