LeetCode 0046 - Permutations
# Hints
STL
DFS
# 题面
Difficulty | Time Complexity Limit | Extra-Memory Complexity Limit |
---|---|---|
Medium |
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]
]
# 题意
给定一个大小为 的不重复集合,求其所有排列。
# 题解
用 STL 的 next_permutation
直接求解即可。
如果禁止使用这一标准库函数,则 DFS 遍历全排列即可。
# AC代码(STL)
class Solution {
public:
vector<vector<int>> permute(vector<int> & nums) {
// 排序
sort(nums.begin(), nums.end());
// 求解
vector<vector<int>> res;
do {
res.push_back(nums);
} while (next_permutation(nums.begin(), nums.end()));
// 返回
return res;
}
};
# AC代码(DFS)
class Solution {
public:
vector<vector<int>> permute(vector<int> & nums) {
// 排序
sort(nums.begin(), nums.end());
// DFS
vector<bool> visited(nums.size(), false);
vector<int> perm(nums.size());
vector<vector<int>> res;
dfs(0, nums, visited, perm, res);
// 返回
return res;
}
// 深度优先搜索
void dfs(int depth, vector<int> & nums, vector<bool> & visited, vector<int> & perm, vector<vector<int>> & res) {
// 递归终止
if (depth == nums.size()) {
res.push_back(perm);
return;
}
// 遍历
for (int i = 0; i < nums.size(); i ++) {
// 递归
if (!visited[i]) {
visited[i] = true;
perm[depth] = nums[i];
dfs(depth + 1, nums, visited, perm, res);
visited[i] = false;
}
}
}
};
- 01
- Reading Papers - Kernel Concurrency06-01
- 02
- Linux Kernel - Source Code Overview05-01
- 03
- Linux Kernel - Per-CPU Storage05-01