leetcode46全排列

    '''
    非递归实现
    index表示生成index+1大小的临时数组作为子数组
    index从1开始，到原始数组的长度
    '''
    if len(nums) <= 1:
        return [nums]

    result = [[nums[0]]]
    index = 1
    while index < len(nums): ###生成index+1大小的子数组，直到等于nums长度终止
        tmp = []
        for perm in result:#遍历每个子数组，每次循环子数组长度+1
            for i in range(index + 1): #i从0到index，
                tmp.append(perm[:i] + [nums[index]] + perm[i:])
                print(perm[:i] + [nums[index]] + perm[i:],i,index)
        result = tmp
        index += 1
    return result

    '''
    python自带的itertools.permutations实现方法
    首先记录123，固定1不动，然后交换2和3，得到132，记录132。
    然后我们将2放到list前，得到213，记录213，然后交换1和3，得到231，记录231。
    然后我们将3放到list前，得到312，记录312，然后交换1和2，得到321，记录321。
    其实看不懂
    '''
    result = list()
    n = len(nums)
    indices = [i for i in range(n)]
    cycles = [i for i in range(n, 0, -1)]
    result.append([nums[i] for i in indices[:n]])
    while n:
        for i in reversed(range(n)):
            cycles[i] -= 1
            if cycles[i] == 0:
                indices[i:] = indices[i + 1:] + indices[i:i + 1]
                cycles[i] = n - i
            else:
                j = cycles[i]
                indices[i], indices[-j] = indices[-j], indices[i]
                result.append([nums[i] for i in indices[:n]])
                break
        else:
            print(result)
            return result

def permute(self, nums):
        '''
        递归实现
        1.递归条件

        '''
        n = len(nums)
        if n == 0 or n == 1:
            return [nums]
        res = []
        for i in range(n):
            list_num = self.permute(nums[0:i]+(nums[i+1:]))
            for j in list_num:
                res.append([nums[i]]+j)
    
        return res
        

参考https://blog.csdn.net/qq_17550379/article/details/82500364