LeetCode Day36 Permutations II
阿新 • • 發佈:2018-12-03
在遞迴函式中要判斷前面一個數和當前的數是否相等,如果相等,前面的數必須已經使用了,即對應的visited中的值為1,當前的數字才能使用,否則需要跳過
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;
}
}
}
}
};
class Solution {
public:
vector<vector<int>> permuteUnique(vector<int>& nums) {
vector<vector<int>> res;
sort(nums.begin(),nums.end());
DFS(nums,res,0);
return res;
}
void DFS(vector<int>& nums,vector<vector<int>> & res,int start){
if(start==nums.size()){res.push_back(nums);return;}
else{
for(int i=start;i<nums.size();i++){
int j = i - 1;
while (j >= start && nums[j] != nums[i]) j--;
if (j == start - 1) {// if nums[i] does not equal to any vals in nums[idx, ... i - 1]
swap(nums[start],nums[i]);
DFS(nums,res,start+1);
swap(nums[start],nums[i]);
}
}
}
}
};