LEETCODE-DAY24-优快云博客

本文链接：https://blog.youkuaiyun.com/qyc_sh/article/details/136762512

文章介绍了如何使用回溯算法解决组合问题，以及针对`nchoosek`问题进行的剪枝优化技巧。通过实例展示了两种解决方案，包括基本的回溯方法和带有剪枝策略的改进版本。

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title: LEETCODE-DAY24
date: 2024-03-15 21:40:07
tags:

今日内容：回溯算法

T1

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        res=[]
        self.backtracking(n,k,1,[],res)
        return res
    def backtracking(self,n,k,idx,path,res):
        if len(path)==k:
            res.append(path.copy())
            return
        for i in range(idx,n+1):
            path.append(i)
            self.backtracking(n,k,i+1,path,res)
            path.pop()

剪枝优化
k-len(path)=n-i+1([i,n]中的个数)

i>n+1-(k-len(path))时,[i,n]中数的个数不足以使path长度到达k，故不需遍历

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

LC example:

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        res = []
        tmp = []
        for i in range(1,k+1):
            tmp.append(i)
        end = []
        for i in range(n-k+1,n+1):
            end.append(i)
        while tmp != end:
            res.append(tmp[:])
            for i in range(1,k+1):
                if tmp[-i] == n + 1 -i:
                    continue
                tmp[k-i] += 1
                for j in range(k-i+1,k):
                    tmp[j] = tmp[j-1] + 1
                break
        res.append(end[:])
        return res