Permutations II 去掉重复的全排列

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]

.

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Backtracking

class Solution {
private:
vector<vector<int> > ret;
public:
void perm(vector<int> &num,int i,int len){
if(i==len){
ret.push_back(num);
return;
}
for(int j=i;j<len;++j){
if(!isSwap(num,i,j))
continue;
swap(num[i],num[j]);
perm(num,i+1,len);
swap(num[j],num[i]);
}
}
bool isSwap(vector<int>& num, int i, int j) {
while (num[i] != num[j] && i < j) i++;
if (i == j) return true;
else return false;
}
vector<vector<int> > permuteUnique(vector<int> &num) {
int len=num.size();
ret.clear();
sort(num.begin(), num.end());
perm(num,0,len);
return ret;
}
};

参考http://blog.csdn.net/morewindows/article/details/7370155