LEETCODE-DAY28-优快云博客

本文链接：https://blog.youkuaiyun.com/qyc_sh/article/details/136877876

本文介绍了如何使用回溯算法解决LeetCode中的两个问题：恢复IP地址（93.题）和子集生成（78.和90.题）。通过`backtracking`函数实现，展示了如何处理子串验证、递归以及处理重复元素的情况。

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title: LEETCODE-DAY28
date: 2024-03-19 14:22:06
tags:

今日内容：93.复原IP地址、78.子集、90.子集II

T1

class Solution:
    def restoreIpAddresses(self, s: str) -> List[str]:
        result = []
        self.backtracking(s, 0, 0, "", result)
        return result

    def backtracking(self, s, start_index, point_num, current, result):
        if point_num == 3:  # 逗点数量为3时，分隔结束
            if self.is_valid(s, start_index, len(s) - 1):  # 判断第四段子字符串是否合法
                current += s[start_index:]  # 添加最后一段子字符串
                result.append(current)
            return

        for i in range(start_index, len(s)):
            if self.is_valid(s, start_index, i):  # 判断 [start_index, i] 这个区间的子串是否合法
                sub = s[start_index:i + 1]
                self.backtracking(s, i + 1, point_num + 1, current + sub + '.', result)
            else:
                break

    def is_valid(self, s, start, end):
        if start > end:
            return False
        if s[start] == '0' and start != end:  # 0开头的数字不合法
            return False
        num = 0
        for i in range(start, end + 1):
            if not s[i].isdigit():  # 遇到非数字字符不合法
                return False
            num = num * 10 + int(s[i])
            if num > 255:  # 如果大于255了不合法
                return False
        return True

T2

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

注意这里不需要return，这并不会导致无限递归，因为每一次递归的下一层i会加1

T3

class Solution:
    def backtracking(self,nums,index,path,res):
        res.append(path.copy())
        for i in range(index,len(path)):
            if i > index and num[i-1]==nums[i]:
                continue
            path.append(nums[i])
            self.backtracking(nums,i+1,path,res)
            path.pop()

    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
        res=[]
        nums.sort()
        self.backtracking(nums,0,[],res)
        return res

输入
nums =
[1,2,2]
输出
[[]]
预期结果
[[],[1],[1,2],[1,2,2],[2],[2,2]]

手抖把backtracking里的len(nums)写成len(path)了

for i in range(index,len(nums)):
            if i > index and num[i-1]==nums[i]:
                continue
            path.append(nums[i])
            self.backtracking(nums,i+1,path,res)
            path.pop()

class Solution:
    def backtracking(self,nums,index,path,res):
        res.append(path.copy())
        for i in range(index,len(nums)):
            if i > index and nums[i-1]==nums[i]:
                continue
            path.append(nums[i])
            self.backtracking(nums,i+1,path,res)
            path.pop()

    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
        res=[]
        nums.sort()
        self.backtracking(nums,0,[],res)
        return res