[LeetCode] Permutations II

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]

.

Solution:

由于数据中有重复，在递归时需要判断是否可以交换。基本的枚举思路是每个数分别出现在第k位，然后排列剩下的元素。这里第k位的元素在枚举过程中需要判断是否已经出现过。

class Solution {
public:
vector<vector<int> > ans;
vector<int> src;
int srcLen;

void perm(int i)
{
if(i == srcLen - 1)
{
vector<int> tmp;
for(int i = 0;i < srcLen;i++)
tmp.push_back(src[i]);
ans.push_back(tmp);
}
else
{
set<int> dic;
for(int k = i;k < srcLen;k++)
{
if(dic.find(src[k]) == dic.end())
{
int t = src[i];
src[i] = src[k];
src[k] = t;
perm(i + 1);
src[k] = src[i];
src[i] = t;
dic.insert(src[k]);
}
}
}
}

vector<vector<int> > permuteUnique(vector<int> &num) {
src = num;
srcLen = num.size();
perm(0);
return ans;
}
};