代码随想录算法训练营第二十四天 | 93.复原IP地址 、78.子集、90.子集II

93.复原IP地址 

照例先写一版很辣鸡但是能AC的：

class Solution:
    def restoreIpAddresses(self, s: str) -> List[str]:
        res = []

        def isValid(a):
            if len(a) > 1 and a[0] == '0':
                return False
            if int(a) > 255:
                return False
            return True

        def dfs(path, start):
            if len(path) == 4 and start >= len(s):
                res.append('.'.join(path))
                return
            
            for end in range(start, len(s)):
                if isValid(s[start: end + 1]):
                    dfs(path + [s[start: end + 1]], end + 1)
        
        dfs([], 0)
        return res

剪枝：

class Solution:
    def restoreIpAddresses(self, s: str) -> List[str]:
        res = []
        
        def isValid(segment):
            if len(segment) > 1 and segment[0] == '0':
                return False
            return 0 <= int(segment) <= 255

        def dfs(path, start):
            if len(path) == 4:
                if start == len(s):
                    res.append('.'.join(path))
                return
            
            for end in range(start + 1, min(start + 4, len(s) + 1)):
                segment = s[start:end]
                if isValid(segment):
                    dfs(path + [segment], end)

        if 4 <= len(s) <= 12:
            dfs([], 0)
        
        return res

78.子集

class Solution:
    def subsets(self, nums: List[int]) -> List[List[int]]:
        res = []
        def dfs(path, start):
            res.append(path[:])
            for i in range(start, len(nums)):
                dfs(path + [nums[i]], i + 1)
        
        dfs([], 0)
        return res

90.子集II

class Solution:
    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        res = []

        def dfs(path, start):
            res.append(path)
            for i in range(start, len(nums)):
                if i > start and nums[i] == nums[i - 1]:
                    continue
                dfs(path + [nums[i]], i + 1)
        
        dfs([], 0)
        return res

后两题比较简单，只给代码了