46. Permutations (Back-Track,Sort)
Given a collection of 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:遍历数组,对于该字母,它可选择与它之后的字母交换或者是不交换=>带回溯的递归
class Solution { public: vector<vector<int> > permute(vector<int> &num) { result.clear(); dfs(num,0); return result; } void dfs(vector<int> num, int depth) { if(depth == num.size()-1) { result.push_back(num); return; } dfs(num,depth+1); int temp = num[depth]; for(int i = depth+1;i< num.size(); i++) { num[depth] = num[i]; num[i] = temp; dfs(num,depth+1); num[i] = num[depth]; num[depth] = temp; } } private: vector<vector<int> > result; };
思路II:
当字符串长度为2时 a1a2 a2a1
当字符串长度为3时 a3a1a2 a1a3a2 a1a2a3 a3a2a1 a2a3a1 a2a1a3
比较可以得到 其实就是把a3(多出来的元素)插在长度为2时的两个字符串的任意位置
时间复杂度:三个for循环 O(n3)
class Solution { public: vector<vector<int>> permute(vector<int>& nums) { int size = nums.size(); int resultSize; int resultIndex; vector<vector<int>> result; vector<int> resultItem(1,nums[0]); result.push_back(resultItem); for(int i = 1; i <size; i++){ //nums[i] is the num to insert resultSize = result.size(); //resultSize in the preceeding insert iterate for(int j = 0; j < resultSize; j++){ //iterate the array to do insertion result[j].push_back(nums[i]); resultIndex = j; for(int k = i-1; k >=0; k--){ //like insertion sort, adjust forward result.push_back(result[resultIndex]); result[result.size()-1][k+1] = result[resultIndex][k]; result[result.size()-1][k] = result[resultIndex][k+1]; resultIndex = result.size()-1; } } } return result; } };