leetcode：Permutations II

 有重复元素的全排序

public class Solution {
//static ArrayList<ArrayList<Integer>> hasUse;
static ArrayList<ArrayList<Integer>> ans;
public ArrayList<ArrayList<Integer>> permuteUnique(int[] num) {
// hasUse = new ArrayList<ArrayList<Integer>>();
ans = new ArrayList<ArrayList<Integer>>();
sort(num,0);
return ans;
}
public void sort(int num[], int cur){
if(cur == num.length){
ArrayList<Integer> temp = new ArrayList<>();
for(int i : num){
temp.add(i);
}
if(ans.size() > 0)
ans.add(ans.size() - 1,temp);
else ans.add(temp);
}
else{
ArrayList<Integer> hasUse = new ArrayList<>();
for(int i = cur; i < num.length; ++i){
if(!hasUse.contains(num[i])){
hasUse.add(num[i]);
int t = num[cur];
num[cur] = num[i];
num[i] = t;
sort(num,cur + 1);
t = num[cur];
num[cur] = num[i];
num[i] = t;
}
}
}
}
}