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Given a collection of numbers that might contain duplicates, return all possible unique permutations.
Example:
Input: [1,1,2] Output: [ [1,1,2], [1,2,1], [2,1,1] ]
这一题和上一题类似,就是数组里有重复的元素。
思路:递归+深度优先搜索
利用标记信息,加一个判断条件,将重复元素去除
class Solution {
public:
vector<vector<int>> permuteUnique(vector<int>& nums) {
vector<vector<int>> ans;
vector<int> cur;
vector<bool> mark(nums.size(), false);
sort(nums.begin(), nums.end());
dfs(nums, 0, ans, cur, mark);
return ans;
}
void dfs(vector<int>& nums, int num, vector<vector<int >>& ans, vector<int>& cur, vector<bool>& mark) {
if (num == nums.size()) {//排完了,存入
ans.push_back(cur);
}
for (int i = 0; i < nums.size(); i++) {
if (mark[i] || i > 0 && nums[i] == nums[i - 1] && !mark[i - 1])
continue;
cur.push_back(nums[i]);
mark[i] = true;
dfs(nums, num + 1, ans, cur, mark);
cur.pop_back();//回溯
mark[i] = false;
}
}
};