【经典题目】【循环DFS】生成数组的全排列

问题描述

给你一个数组，生成这个数组中元素的全排列。

思路

经典的循环dfs。要点是我们需要设置visited数组来指代其是否被遍历过。

代码

class Solution:
    def islandPerimeter(self, grid):
        if not grid:
            return []
        visited = [False for i in grid]
        res = []
        
        def dfs(last):
            if len(last) == len(grid):
                res.append(last)
            for i in range(len(grid)):
                if not visited[i]:
                    last += grid[i]
                    visited[i] = True
                    dfs(last)
                    visited[i] = False
                    last = last[:-1]
        dfs('')
        return res

if __name__ == '__main__':
    s = Solution()
    print(s.islandPerimeter(['1', '2', '3', '4','5']).__len__())

经典变型：如果我们改一下题目，不生成全排列，而是固定长度的排列，那么我们只需要改一下last长度的判断条件即可。

class Solution:
    def islandPerimeter(self, grid, length):
        if not grid:
            return []
        visited = [False for i in grid]
        res = []

        def dfs(last):
            if len(last) == length:
                res.append(last)
            for i in range(len(grid)):
                if not visited[i]:
                    last += grid[i]
                    visited[i] = True
                    dfs(last)
                    visited[i] = False
                    last = last[:-1]
        dfs('')
        return res

if __name__ == '__main__':
    s = Solution()
    print(s.islandPerimeter(['1', '2', '3', '4','5'], 3))