2026.1.29 复原ip地址 & 子集 & 子集2

多打代码2026-01-30 8:37

python 复制代码

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

        def backtrack(startIndx : int) -> None:
            if startIndx == n and len(path) == 4:
                res.append(".".join(path))
                return
            remain_char = len(s) - startIndx
            remain_seg = 4 - len(path)
            if remain_char > 3 * remain_seg or remain_char < remain_seg :
                # 剪枝：如果剩余的字符串太多或太少，直接返回
                return

            for i in range(startIndx, min(startIndx + 3, len(s))):
                # 每个被分割的字串都在3或者字符串长度结尾以内
                subStr = s[startIndx:i + 1]

                if len(subStr) > 1 and subStr[0] == '0':
                    # 不能有前导零
                    break
                if int(subStr) > 255:
                    # 数值不大于255
                    break
                
                path.append(subStr)
                backtrack(i + 1)
                path.pop()
        backtrack(0)
        return res

78. 子集

这里用回溯和dp都能做，在实际算法中，回溯更偏向于搜索所有路径，dp更偏向于找出最优解

回溯

python 复制代码

class Solution:
    def subsets(self, nums: List[int]) -> List[List[int]]:
        # 决策树，每次都可以选择拿或不拿当前元素
        ans = []
        path = []
        n = len(nums)
        def backtrack(startIdx:int) -> None:
            ans.append(path[:])
            for i in range(startIdx, n):
                # 将当前元素放进子集中，代表选择当前元素
                path.append(nums[i])
                # 并进行下一次迭代，从剩余的列表中找出包含当前nums[i]的元素
                backtrack(i + 1)
                # 弹出当前元素，代表不选，下一次循环接着从下一个元素开始选
                path.pop()
            
        backtrack(0)
        return ans

python 复制代码

class Solution:
    def subsets(self, nums: List[int]) -> List[List[int]]:
        dp = [[]]

        for num in nums:
            new_subsets = []
            for cur in dp:
                new_subsets.append(cur + [num])
            
            dp.extend(new_subsets)
        
        return dp

https://qcn7yki7og1q.feishu.cn/record/OWgEretm4eHCFec2E8GcYxrPnv9

杀死同一树层的判断逻辑

if i > startIdx and nums $i$ == nums $i - 1$

python 复制代码

class Solution:
    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
        ans = []
        path = []
        n = len(nums)
        nums.sort()
        def backstrack(startIdx:int) -> None:
            ans.append(path[:])

            for i in range(startIdx, n):
                if i > startIdx and nums[i] == nums[i - 1]:
                    continue
                path.append(nums[i])
                backstrack(i + 1)
                path.pop()

        backstrack(0)
        return ans