# LeetCode46. Permutations（递归 / 回溯）

## LeetCode46. Permutations（递归 / 回溯）

• Medium
• Accepted：359,004
• Submissions：661,126

Given a collection of distinct integers, return all possible permutations.

Example:

Input: [1,2,3]
Output:
[
[1,2,3],
[1,3,2],
[2,1,3],
[2,3,1],
[3,1,2],
[3,2,1]
]

### 链接

https://leetcode.com/problems/permutations

### 代码

class Solution {
private:
vector<vector<int> > ans; // 保存答案
vector<bool> vis; // 判断是否已经使用
void solve(const vector <int> & nums, int index, vector<int> & p)
{
if (index == nums.size())
{
ans.push_back(p);
return ;
}
for (int i = 0; i < nums.size(); ++i)
{
if (!vis[i])
{
p.push_back(nums[i]);
vis[i] = true;
solve(nums, index + 1, p);
p.pop_back(); // 回溯
vis[i] = false;
}
}
}
public:
vector<vector<int>> permute(vector<int>& nums) {
ans.clear();
if (nums.empty())
return ans;
vector <int> p;
vis = vector<bool> (nums.size(), false);
solve(nums, 0, p);
return ans;
}
};
• Runtime: 3 ms, faster than 79.22% of C++ online submissions for Permutations.
• Memory Usage: 7.7 MB, less than 56.70% of C++ online submissions for Permutations.
class Solution
{
public:
vector<vector<int>> permute(vector<int> &nums)
{
result.clear();
n = nums.size();
if (n == 0)
return result;
getSets(nums, 0);
return result;
}
private:
int n;
vector<vector<int>> result;

void getSets(vector<int> &nums, int idx)
{
if (idx == n)
{
result.push_back(nums);
return ;
}
else if (idx < n)
{
// 交换位置，每次有 n - idx 个选择
for (int i = idx; i < n; ++i)
{
swap(nums[i], nums[idx]);
getSets(nums, idx + 1);
swap(nums[i], nums[idx]);
}
}
}
};