216.组合总和III

相较于组合问题多了一个剪枝操作：当前总和大于目标总和时候，直接剪枝

class Solution:
    def combinationSum3(self, k: int, n: int) -> List[List[int]]:
        result = []  # 存放结果集
        self.backtracking(n, k, 0, 1, [], result)
        return result

    def backtracking(self, targetSum, k, currentSum, startIndex, path, result):
        if currentSum > targetSum:  # 剪枝操作
            return  # 如果path的长度等于k但currentSum不等于targetSum，则直接返回
        if len(path) == k:
            if currentSum == targetSum:
                result.append(path[:])
            return
        for i in range(startIndex, 9 - (k - len(path)) + 2):  # 剪枝
            currentSum += i  # 处理
            path.append(i)  # 处理
            self.backtracking(targetSum, k, currentSum, i + 1, path, result)  
            currentSum -= i  # 回溯
            path.pop()  # 回溯

17.电话号码的字母组合

直接对着模板写就行，注意处理字符串的特殊方式

class Solution:
    def __init__(self):
        self.letterMap = [
            "",     # 0
            "",     # 1
            "abc",  # 2
            "def",  # 3
            "ghi",  # 4
            "jkl",  # 5
            "mno",  # 6
            "pqrs", # 7
            "tuv",  # 8
            "wxyz"  # 9
        ]
        self.result = []
        self.s = ""
    
    def backtracking(self, digits, index):
        if index == len(digits):
            self.result.append(self.s)
            return
        digit = int(digits[index])    # 将索引处的数字转换为整数
        letters = self.letterMap[digit]    # 获取对应的字符集
        for i in range(len(letters)):
            self.s += letters[i]    # 处理字符
            self.backtracking(digits, index + 1)    # 递归调用，注意索引加1，处理下一个数字
            self.s = self.s[:-1]    # 回溯，删除最后添加的字符
    
    def letterCombinations(self, digits):
        if len(digits) == 0:
            return self.result
        self.backtracking(digits, 0)
        return self.result