代码随想录算法训练营第二十五日| LC216.组合总和III LC17.电话号码的字母组合-优快云博客

本文链接：https://blog.youkuaiyun.com/EdisonW8/article/details/127960923

本文介绍了两种典型的回溯算法应用：组合总和III问题，寻找所有可能的组合使数字之和等于目标值；电话号码的字母组合问题，将数字转换为对应的字母组合。通过具体代码实现展示了回溯算法的设计思路。

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LC216.组合总和III:

class Solution:
    def __init__(self):
        self.res = []
        self.sum = 0
        self.path = []
    
    def combinationSum3(self, k: int, n: int) -> List[List[int]]:
        self.backtracking(k, n, 1)
        return self.res

    def backtracking(self, k: int, n: int, start_num: int):
        if self.sum > n: # 剪枝
            return
        if len(self.path) == k:  # len(path)==k时不管sum是否等于n都会返回
            if self.sum == n:
                self.res.append(self.path[:])
            return
        for i in range(start_num, 10 - (k - len(self.path)) + 1):
            self.path.append(i)
            self.sum += i
            self.backtracking(k, n, i+1)
            self.path.pop()
            self.sum -= i

LC17.电话号码的字母组合

class Solution:
    def __init__(self):
        self.answers: List[str] = []
        self.answer: str = ''
        self.letter_map = {
            '2': 'abc',
            '3': 'def',
            '4': 'ghi',
            '5': 'jkl',
            '6': 'mno',
            '7': 'pqrs',
            '8': 'tuv',
            '9': 'wxyz'
        }

    def letterCombinations(self, digits: str) -> List[str]:
        self.answers.clear()
        if not digits: return []
        self.backtracking(digits, 0)
        return self.answers
    
    def backtracking(self, digits: str, index: int) -> None:
        # 回溯函数没有返回值
        # Base Case
        if index == len(digits):    # 当遍历穷尽后的下一层时
            self.answers.append(self.answer)
            return 
        # 单层递归逻辑  
        letters: str = self.letter_map[digits[index]]
        for letter in letters:
            self.answer += letter   # 处理
            self.backtracking(digits, index + 1)    # 递归至下一层
            self.answer = self.answer[:-1]  # 回溯