[LeetCode]046-Permutations
2016-05-06 22:35
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题目:
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].
Solutions:
思路1:
比较容易想到,固定住第一位的数,然后全排列。运用递归的思想,每次递归都是固定住第一位。
代码:
思路2:
这个思路比较常见,依次交换,将每一位的数字依次与后面位置的数字进行交换。
代码:
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].
Solutions:
思路1:
比较容易想到,固定住第一位的数,然后全排列。运用递归的思想,每次递归都是固定住第一位。
代码:
vector<vector<int>> permute(vector<int>& nums) { int N = nums.size(); vector<vector<int>> result; if(N == 1) { result.push_back(nums); return result; } vector<int> temp; //暂存余下的nums vector<vector<int>> m_temps; for(int i = 0;i < N;i++) { temp = nums; temp.erase(temp.begin()+i); //删除下标为i的数,然后进行余下的全排列,这个数插入到开头 m_temps = permute(temp); //余下的全排列 for(int j = 0;j < m_temps.size();j ++) { vector<int> s = m_temps[j]; //获取每一个全排列 s.insert(s.begin(),nums[i]); //将删除的数插入到初始位置 result.push_back(s); } } return result; }
思路2:
这个思路比较常见,依次交换,将每一位的数字依次与后面位置的数字进行交换。
代码:
void perm(vector<vector<int>>& result,vector<int> nums,int i) { if(i == nums.size()-1) result.push_back(nums); for(int j = i ;j < nums.size();j++) { swap(nums[i],nums[j]); perm(result,nums,i+1); swap(nums[i],nums[j]); } } vector<vector<int>> permute(vector<int>& nums) { int N = nums.size(); vector<vector<int>> result; perm(result,nums,0); return result; }
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