标签:lap splay ica display cli alt uniq view nta
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
Example:
Input: [1,1,2]
Output:
[
[1,1,2],
[1,2,1],
[2,1,1]
]
1 class Solution { 2 public: 3 vector<vector<int>> permuteUnique(vector<int>& nums) { 4 set<vector<int>>ans; 5 ans.insert(vector<int>(1, nums[0])); 6 for (int i = 1; i < nums.size(); i++) { 7 set<vector<int>>tmp = ans; 8 ans.clear(); 9 for (auto j = tmp.begin(); j != tmp.end(); j++) { 10 vector<int>now = *j; 11 for (int k = 0; k <= now.size(); k++) { 12 now.insert(now.begin() + k, nums[i]); 13 ans.insert(now); 14 now.erase(now.begin() + k); 15 } 16 } 17 } 18 return vector<vector<int>>(ans.begin(),ans.end()); 19 } 20 };
单纯只是把上一道改了一下
19.2.7 [LeetCode 47] Permutations II
标签:lap splay ica display cli alt uniq view nta
原文地址:https://www.cnblogs.com/yalphait/p/10354993.html