Thursday, March 18, 2021, the weather is fine [Do not lament the past, do not waste the present, do not fear the future]
Contents of this article
1. Introduction
2. Problem solution (retrospecting Dafa)
2.1 Temporary array method
class Solution {
public:
unordered_set<int> uset; // 利用哈希表判断元素是否遍历过了
vector<vector<int>> res;
vector<vector<int>> permute(vector<int>& nums) {
vector<int> tmp; // 临时数组，存放每一个排列的结果
dfs(nums,tmp,0);
return res;
}
void dfs(vector<int>& nums, vector<int>& tmp, int n){
// 终止条件，所有位都遍历完了
if(n==nums.size()){
res.push_back(tmp);
return;
}
for(int num:nums){
// 如果 num 还没有遍历
if(uset.find(num)==uset.end()){
// 加入哈希表和临时数组
uset.insert(num);
tmp.push_back(num);
// 继续遍历下一位
dfs(nums,tmp,n+1);
// 遍历完之后记得还原，避免分支污染
uset.erase(num);
tmp.pop_back();
}
}
}
};
This is the answer I wrote when I first did it, and it's easier to understand. Although it can be passed, it is more time-consuming than the official method (the exchange method below).
2.2 Exchange method
class Solution {
public:
void backtrack(vector<vector<int>>& res, vector<int>& output, int first, int len){
// 所有数都填完了
if (first == len) {
res.emplace_back(output);
return;
}
for (int i = first; i < len; ++i) {
// 动态维护数组
swap(output[i], output[first]);
// 继续递归填下一个数
backtrack(res, output, first + 1, len);
// 撤销操作
swap(output[i], output[first]);
}
}
vector<vector<int>> permute(vector<int>& nums) {
vector<vector<int> > res;
backtrack(res, nums, 0, (int)nums.size());
return res;
}
};
The exchange method is also a standard method to solve the full arrangement problem. Although the code is simple, it takes a little effort to understand it.
The following picture helps to understand:
*Complexity analysis
The calculation of time complexity is more difficult to understand. You can refer to the following picture to understand:
add up the number of exchanges of each layer, you can get the total number of calls:
Then you can derive the above formula.
references
https://leetcode-cn.com/problems/permutations/solution/quan-pai-lie-by-leetcode-solution-2/
https://blog.csdn.net/u013905744/article/details/113779407