LeetCode 47.Permutations II

题目：

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]

.
分析与解答：

这个题比上个题目更难了，问题的关键在于如何判断重复。我自己的方法是先排序，然后遇到重复进栈的数就跳过。

class Solution {
public:
vector<vector<int> > result;
vector<vector<int> > permuteUnique(vector<int> &num) {
sort(num.begin(),num.end());
vector<int> tempresult;
vector<int> path;//存放选过数的下标
dfs(num,tempresult,path);
return result;
}
void dfs(vector<int> &num,vector<int> &tempresult,vector<int> &path){
if(tempresult.size() == num.size()){
result.push_back(tempresult);
return;
}
int last = INT_MAX;
for(int i = 0;i < num.size();++i){
auto index = find(path.begin(),path.end(),i);
if(index == path.end() && num[i] != last){//这个数没被选过
tempresult.push_back(num[i]);
path.push_back(i);
dfs(num,tempresult,path);
tempresult.pop_back();
path.pop_back();
last = num[i];
}

}
}
};

网上有一个更简洁的版本：

class Solution {
public:
void recursion(vector<int> num, int i, int j, vector<vector<int> > &res) {
if (i == j-1) {
res.push_back(num);
return;
}
for (int k = i; k < j; k++) {
if (i != k && num[i] == num[k]) continue;
swap(num[i], num[k]);
recursion(num, i+1, j, res);
}
}
vector<vector<int> > permuteUnique(vector<int> &num) {
sort(num.begin(), num.end());
vector<vector<int> >res;
recursion(num, 0, num.size(), res);
return res;
}
};