【leetcode】Permutations (middle)
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]
.
求没有重复数字的全排列
思路:用的标准回溯法,一次AC
class Solution { public: vector<vector<int> > permute(vector<int> &num) { vector<vector<int>> ans; if(num.empty()) { return ans; } vector<int> X(num.size()); vector<vector<int>> S(num.size()); int k = 0; S[k] = num; while(k >= 0) { while(!S[k].empty()) { X[k] = S[k].back(); S[k].pop_back(); if(k < num.size() - 1) { k++; S[k] = num; for(int i = 0; i < k; i++) { vector<int>::iterator it; if((it = find(S[k].begin(), S[k].end(), X[i])) != S[k].end()) { S[k].erase(it); } } } else { ans.push_back(X); } } k--; } return ans; } };
网上也有用递归的
class Solution { public: vector<vector<int> > permute(vector<int> &num) { vector<vector<int> > result; permuteRecursive(num, 0, result); return result; } // permute num[begin..end] // invariant: num[0..begin-1] have been fixed/permuted void permuteRecursive(vector<int> &num, int begin, vector<vector<int> > &result) { if (begin >= num.size()) { // one permutation instance result.push_back(num); return; } for (int i = begin; i < num.size(); i++) { swap(num[begin], num[i]); permuteRecursive(num, begin + 1, result); // reset swap(num[begin], num[i]); } } };