leecode

leecode20，有效的括号，栈

class Solution:
    def isValid(self, s: str) -> bool:
        def check(ch1,ch2):
            if ch1 == '[' and ch2 == ']':
                return True
            elif ch1 == '(' and ch2 == ')':
                return True
            elif ch1 == '{' and ch2 == '}':
                return True
            else:
                return False
        stack = []
        for i in range(len(s)):
            if len(stack) == 0 or check(stack[-1],s[i]) == False:
                stack.append(s[i])
            elif check(stack[-1],s[i]) == True:
                stack.pop()
        if len(stack) == 0:
            return True
        else:
            return False

leecode22，括号生成，dfs+回溯，注意dfs时候的判断条件

class Solution:
    def generateParenthesis(self, n: int):
        ans = []
        path = []
        def dfs(l,r,path):
            if r==n:
                ans.append("".join(path))
                return
            if lr:
                path.append(')')
                dfs(l,r+1,path)
                path.pop()
            return 
        dfs(0,0,[])
        return ans

leecod17，电话号码的组合，dfs+回溯:停止条件，循环回溯剪枝

class Solution:
    def letterCombinations(self, digits: str):
        ans = []
        res = []
        hash_map = {2:"abc",3:"def",4:"ghi",5:"jkl",6:"mno",7:"pqrs",8:"tuv",9:"wxyz"}
        if digits == "":
            return []
        def dfs(i,path):
            if i == len(digits):
                ans.append("".join(path))
                return
            for ch in hash_map[int(digits[i])]:
                path.append(ch)
                dfs(i+1,path)
                path.pop()
            return
        dfs(0,[])
        return ans

17，20，22，39