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],
[2,1,1]
]
46. Permutations 的拓展，这题数组含有重复的元素。解法和46题，主要是多出处理重复的数字。
先对nums排序，使得相同的元素排在一起。新建一个大小与nums相同visited数组，用来标记在本次DFS读取中，位置i的元素是否已经被添加。
如果当前的数与前一个数相等，并且前一个数未被添加到list中，则可跳过这个数。
C++:
class Solution {
public:
vector<vector<int> > permuteUnique(vector<int> &num) {
vector<vector<int> > res;
vector<int> out;
vector<int> visited(num.size(), 0);
sort(num.begin(), num.end());
permuteUniqueDFS(num, 0, visited, out, res);
return res;
}
void permuteUniqueDFS(vector<int> &num, int level, vector<int> &visited, vector<int> &out, vector<vector<int> > &res) {
if (level >= num.size()) res.push_back(out);
else {
for (int i = 0; i < num.size(); ++i) {
if (visited[i] == 0) {
if (i > 0 && num[i] == num[i - 1] && visited[i - 1] == 0) continue;
visited[i] = 1;
out.push_back(num[i]);
permuteUniqueDFS(num, level + 1, visited, out, res);
out.pop_back();
visited[i] = 0;
}
}
}
}
};
类似题目：
[LeetCode] 46. Permutations 全排列
[LeetCode] 31. Next Permutation 下一个排列
[LeetCode] 60. Permutation Sequence 序列排序