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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].
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Analysis:
We sort the array first, then we use recursive method to construct the solution. At each position, it will place each current available number on it. For handling duplicates, if we have already put a num X on the current place, we then skip all X in the following iterations.
Solution:
1 public class Solution { 2 public List<List<Integer>> permuteUnique(int[] num) { 3 boolean[] used = new boolean[num.length]; 4 Arrays.fill(used,false); 5 Arrays.sort(num); 6 List<List<Integer>> resSet = new ArrayList<List<Integer>>(); 7 List<Integer> curStr = new ArrayList<Integer>(); 8 permuteRecur(num,used,0,curStr,resSet); 9 return resSet; 10 }11 12 public void permuteRecur(int[] num, boolean[] used, int cur, List<Integer> curStr, List<List<Integer>> resSet){13 if (cur==num.length){14 List<Integer> res = new ArrayList<Integer>();15 res.addAll(curStr);16 resSet.add(res);17 return;18 }19 20 boolean init = true;21 int lastVal = -1;22 for (int i=0;i<num.length;i++)23 if (used[i] || (!init && num[i]==lastVal))24 continue;25 else {26 init = false;27 curStr.add(num[i]);28 used[i]=true;29 permuteRecur(num,used,cur+1,curStr,resSet);30 used[i]=false;31 curStr.remove(curStr.size()-1);32 lastVal = num[i];33 }34 }35 }
Leetcode-Permutations II
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