Permutations
Given a collection of distinct numbers, return all possible permutations.
For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].
思路:DFS
k∈[0, size-1],k位置之前的元素已经排好,需要全排列[k,size-1]的所有元素。回溯的时候需要恢复nums,本方法生成的序列没有按照字典序。
class Solution {
public:
vector<vector<int> > permute(vector<int>& nums) {
vector<vector<int> > res;
if(nums.size() == 0) return res;
helper(nums, res, 0);
return res;
}
private:
void helper(vector<int>& nums, vector<vector<int> >& res, int k){
if(k == nums.size()){
res.push_back(nums);
return;
}
for(int i = k; i < nums.size(); i++){
swap(nums[k], nums[i]);
helper(nums, res, k+1);
swap(nums[k], nums[i]);
}
}
};
Permutations II
记录一下每个位置k
使用的元素不重复就行了
#include<unordered_set>
class Solution {
public:
void helper(vector<int>& nums, int k){
if(k == nums.size()-1) res.push_back(nums);
else{
unordered_set<int> s;
for(int i = k; i < nums.size(); ++i){
if(s.find(nums[i]) != s.end()) continue;
s.insert(nums[i]);
swap(nums[k], nums[i]);
helper(nums, k+1);
swap(nums[k], nums[i]);
}
}
}
vector<vector<int>> permuteUnique(vector<int>& nums) {
res = vector<vector<int> >();
helper(nums, 0);
return res;
}
private:
vector<vector<int> > res;
};