Leetcode 刷题笔记1 回溯算法part01

leetcode 77 组合

回溯算法的入门，回溯算法更像是一种枚举的策略，本题的组合优化过程可以理解。通过pycharm的调试，对回溯有了初步认识

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        result = []
        self.backtracking(n, k, 1, [], result)
        return result
    def backtracking(self, n, k , startIndex, path, result):
        if len(path) == k:
            result.append(path[:])
            return
        # 优化后for i in range(startIndex,  n - (k - len(path)) + 2)
        for i in range(startIndex, n + 1):
            path.append(i)
            self.backtracking(n, k , i + 1, path, result)
            path.pop()

leetcode 216 组合总和|||

这题是上题的改版属于基础，值得注意的是回溯算法的剪枝操作

class Solution:
    def combinationSum3(self, k: int, n: int) -> List[List[int]]:
        result = []
        self.backtracking(n, k, 0,)
        return result
    def backtracking(self, targetSum, k, currentSum, startIndex, path, result):
        if currentSum > targetSum:
            return
        if len(path) == k:
            if currentSum == targetSum:
                result.append(path[:])
            return
        for i in range(startIndex, 9 - (k - len(path)) + 2):
            currentSum += i
            path.append(i)
            self.backtracking(targetSum, k, currentSum, i + 1, path, result):
            currentSum -= i
            path.pop()

leetcode 17 电话号码的字母组合

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        self.letterMap = [
            "",
            "",
            "abc",
            "def",
            "ghi",
            "jkl",
            "mno",
            "pqrs",
            "tuv",
            "wxyz"
        ]
        self.result = []
        self.s = ""
        if len(digits) == 0:
            return self.result
        self.backtracking(digits, 0)
        return self.result
    def backtracking(self, digits, index):
        if index == len(digits):
            self.result.append(self.s)
            return 
        digit = int(digits[index])
        letters = self.letterMap[digit]
        for i in range(len(letters)):
            self.s += letters[i]
            self.backtracking(digits, index + 1)
            self.s = self.s[:-1]