Leetcode---Permutations II
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Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example,[1,1,2]
have the following unique permutations:[1,1,2]
, [1,2,1]
, and [2,1,1]
.
Have you met this question in a real interview?
DFS,并且去掉这样的情况:
某个数和上一个数相等,但是上一个数却没有被遍历到---->也就是说,结果中改变了源中数的顺序。
vector<int> temp;unordered_set<int> m;vector<vector<int> > result;void dfs(int level,vector<int> &num){ if(level==num.size()){ result.push_back(temp); return; } for(int i=0;i<num.size();i++){ if(m.find(i)!=m.end()) continue; if(i>0 && num[i]==num[i-1] && m.find(i-1)==m.end()) continue; temp.push_back(num[i]); m.insert(i); dfs(level+1,num); temp.pop_back(); m.erase(i); }}vector<vector<int> > permuteUnique(vector<int> &num) { sort(num.begin(),num.end()); dfs(0,num); return result;}
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