2019年8月19日——leetcode

最新推荐文章于 2023-11-10 17:33:48 发布

技术宅zch

最新推荐文章于 2023-11-10 17:33:48 发布

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分类专栏：基础算法

本文链接：https://blog.youkuaiyun.com/qq_19672707/article/details/99772436

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13 篇文章

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本文介绍了三道LeetCode的编程题目，包括如何生成所有可能的满二叉树，求解最长有效括号的解决方案，以及找出字符串中无重复字符的最长子串的方法。链接提供了详细题目描述和解答思路。

894.所有可能的满二叉树

https://leetcode-cn.com/problems/all-possible-full-binary-trees/

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    memo = {0: [],1:[TreeNode(0)]}
    
    def allPossibleFBT(self, N: int) -> List[TreeNode]:
        if N not in Solution.memo:
            ans = []
            for x in range(N):
                y = N - 1 -x
                for left in self.allPossibleFBT(x):
                    for right in self.allPossibleFBT(y):
                        bns = TreeNode(0)
                        bns.left = left
                        bns.right = right
                        ans.append(bns)
            Solution.memo[N]=ans
        return Solution.memo[N]

32.最长有效括号

https://leetcode-cn.com/problems/longest-valid-parentheses/

class Solution:
    def longestValidParentheses(self, s: str) -> int:
        ans = 0
        dp = [0] * len(s)
        for i in range(1,len(s)):
            if s[i] == ")":
                if s[i-1] == "(":
                    dp[i] = (dp[i-2] if i >= 2 else 0) + 2
                elif i -  dp[i-1] > 0 and s[i-dp[i-1]-1] == "(":
                    dp[i] = dp[i-1] + (dp[i - dp[i-1] - 2] if i - dp[i-1] >= 2 else 0) + 2
                ans = max(ans,dp[i])
                
        return ans

3.无重复的最长子串

https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        ans = 0
        i = 0   # 第一个指针
        char_dict = {}
        for j in range(len(s)):
            if s[j] in char_dict.keys():
                i = max(char_dict[s[j]],i)
            ans = max(ans, j- i + 1)
            char_dict[s[j]] = j+1
        return ans