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].
DFS + 剪枝
class Solution {
public:
void DFS(vector<int> &num, int size,vector<int> temp)
{
if(size == n){
result.push_back(temp);
return ;
}
for(int i = ; i< n; i++)
{
if(flag[i] || (i!= &&flag[i-] && num[i] == num[i-] ) )
continue;
temp[size] = num[i];
flag[i] = true;
DFS(num, size+, temp);
flag[i] = false;
}
}
vector<vector<int> > permuteUnique(vector<int> &num) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
n = num.size();
result.clear();
sort(num.begin(), num.end());
if(n == ) return result;
flag.resize(n,false);
vector<int> temp(n,) ;
DFS(num,,temp);
return result ;
}
private :
int n;
vector<bool> flag;
vector<vector<int>> result;
};
这里解释下剪枝的原理: 有重复元素的时候,因为重复的元素处理无序所以导致重复,所以只要给重复的元素进入temp定义一个次序就可以去掉重复。这里定义次序的规则是: 先对所有元素排序对于有重复的元素,必须是排在后面的元素比排在前面的元素先进入temp
重写后,貌似比第一个版本要快一点
class Solution {
public:
void DFS(vector<int> &num, vector<int> &tp, vector<bool> flag)
{
if(num.size() == tp.size()){
res.push_back(tp);
return;
}
for(int i = 0; i< num.size(); i++)
{
if(flag[i] == true) continue;
if(i != 0 && num[i] == num[i-1] && flag[i-1] == false) continue;
flag[i] = true;
tp.push_back(num[i]);
DFS(num, tp, flag);
tp.pop_back();
flag[i] = false;
}
}
vector<vector<int> > permuteUnique(vector<int> &num) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
res.clear();
int len = num.size();
if(len < 1) return res;
sort(num.begin(), num.end());
vector<bool> flag(len, false);
vector<int> tp;
DFS(num, tp, flag);
return res;
}
private:
vector<vector<int>> res;
};
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