[leetcode] （Dec,23）每天两道题目，配 python参考解答

Afan

题目（一）：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.




题目（二）：Longest Valid Parentheses
Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring.
For "(()", the longest valid parentheses substring is "()", which has length = 2.
Another example is ")()())", where the longest valid parentheses substring is "()()", which has length = 4.


图片割开参考答案：

题目（一）：                                                      题目（二）：







python参考解答：
题目（一）：Count and Say
[code]class Solution:
    # @return a string
    def count(self,s):
        t=''; count=0; curr='#'
        for i in s:
            if i!=curr:
                if curr!='#':
                    t+=str(count)+curr
                curr=i
                count=1
            else:
                count+=1
        t+=str(count)+curr
        return t
    def countAndSay(self, n):
        s='1'
        for i in range(2,n+1):
            s=self.count(s)
        return s[/code]
题目（二）：Longest Valid Parentheses
[code]class Solution:
    # @param s, a string
    # @return an integer
    def longestValidParentheses(self, s):
        maxlen = 0
        stack = []
        last = -1
        for i in range(len(s)):
            if s【i】=='(':
                stack.append(i)     # push the INDEX into the stack!!!!
            else:
                if stack == []:
                    last = i
                else:
                    stack.pop()
                    if stack == []:
                        maxlen = max(maxlen, i-last)
                    else:
                        maxlen = max(maxlen, i-stack[len(stack)-1])
        return maxlen[/code]

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pc_coder

顶了再看，谢谢楼主