Leetcode 刷题笔记1 回溯算法part02

leetcode 39 组合总和

因为此题的题干要求可以不限元素的使用次数，因此需要调整startIndex

class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        result = []
        self.backtracking(candidates, target, 0, 0, [], result)
        return result

    def backtracking(self, candidates, target, total, startIndex, path, result):
        if total > target:
            return
        if total == target:
            result.append(path[:])
            return
        for i in range(startIndex, len(candidates)):
            total += candidates[i]
            path.append(candidates[i])
            self.backtracking(candidates, target, total, i, path, result)
            total -= candidates[i]
            path.pop()

leetcode 40 组合总和||

这题的关键是去重，暴力解法是想使用字典。正解是先对数组排序， 然后比较candiates[i] 和candiates[i-1] 在i > startIndex时。

class Solution:
    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        result = []
        candidates.sort()
        self.backtracking(candidates, target, 0, 0, [], result)
        return result
    def backtracking(self, candidates, target, total, startIndex, path, result):
        if total == target:
            result.append(path[:])
            return
        for i in range(startIndex, len(candidates)):
            # 去重操作
            if startIndex < i and candidates[i] == candidates[i - 1]:
                continue 
            if total > target:
                break 
            total += candidates[i]
            path.append(candidates[i])
            self.backtracking(candidates, target, total, i + 1, path, result)
            total -= candidates[i]
            path.pop()

leetcode 131 分割回文串

终止条件这里刚开始没搞懂，

class Solution:
    def partition(self, s: str) -> List[List[str]]:
        result = []
        self.backtracking(s, 0, [], result)
        return result
    def backtracking(self, s, startIndex, path, result):
        if startIndex == len(s):
            result.append(path[:])
            return
        for i in range(startIndex, len(s)):
            if s[startIndex: i + 1] == s[startIndex: i + 1][::-1]:
                path.append(s[startIndex: i + 1])
                self.backtracking(s, i + 1, path, result)
                path.pop()