题解 | #没有重复项数字的全排列#

回溯的思想，当遍历到响应的元素后，注意还有一步测回的操作。

class Solution:

    def permute(self , num: List[int]) -> List[List[int]]:
        # write code here
        if not num:
            return None
        num.sort()
        res = []
        self.backtrack(num, res, 0)
        return res
        
        
    def backtrack(self, num, res, index):
        if index == len(num) - 1:
            res.append(num)
        else:
            for i in range(index, len(num)):
                temp = num[i]
                num[i] = num[index]
                num[index] = temp 
                self.backtrack(num, res, index + 1)
                temp = num[i]
                num[i] = num[index]
                num[index] = temp