1. 题目信息

给定一个没有重复数字的序列，返回其所有可能的全排列。

示例:输入: [1,2,3]
输出:
[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]
]

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/permutations
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2. 解题

类似题目：
LeetCode 996. 正方形数组的数目（回溯+剪枝）

2.1 利用hash map解决

• 在hash map中查找不到的元素，将其push进数组
• 递归处理
  

class Solution {
public:vector<vector<int>> permute(vector<int>& nums) {vector<int> row;vector<vector<int>> ans;unordered_map<int, int> m;bt(nums, 0, m, row, ans);return ans;}void bt(vector<int>& nums, int count, unordered_map<int, int> m, vector<int> row, vector<vector<int>>& ans){if(count == nums.size()){ans.push_back(row);   return;}for(int i = 0; i < nums.size(); ++i){if(m.find(i) == m.end()){m[i] = 1;row.push_back(nums[i]);bt(nums,count+1,m,row,ans);row.pop_back();m.erase(i);}}}
};

2.2 改用bool数组判断是否出现过

用bool数组做判断，节省内存

class Solution {
public:vector<vector<int>> permute(vector<int>& nums) {vector<int> row;vector<vector<int>> ans;bool occurred[nums.size()];memset(occurred, 0, sizeof(bool)*nums.size());bt(nums, 0, occurred, row, ans);return ans;}void bt(vector<int>& nums, int count, bool *occurred, vector<int> row, vector<vector<int>>& ans){if(count == nums.size()){ans.push_back(row);   return;}for(int i = 0; i < nums.size(); ++i){if(occurred[i] == false){occurred[i] = true;row.push_back(nums[i]);bt(nums,count+1,occurred,row,ans);row.pop_back();occurred[i] = false;}}}
};

class Solution:def permute(self, nums: List[int]) -> List[List[int]]:import itertoolsreturn list(itertools.permutations(nums, len(nums)))

32 ms 15.1 MB Python3