Change the way handling duplicates.

PermutationsII.h

@@ -1,7 +1,7 @@
 /*
  Author:     Annie Kim, anniekim.pku@gmail.com
  Date:       Apr 29, 2013
- Update:     Jul 19, 2013
+ Update:     Sep 2, 2013
  Problem:    Permutations II
  Difficulty: Easy
  Source:     http://leetcode.com/onlinejudge#question_47
@@ -11,39 +11,41 @@
  [1,1,2] have the following unique permutations:
  [1,1,2], [1,2,1], and [2,1,1].
 
- Solution: dfs...use <set> to avoid duplicates.
+ Solution: dfs...
  */
 
 class Solution {
 public:
     vector<vector<int>> res;
     vector<vector<int>> permuteUnique(vector<int> &num) {
         res.clear();
-        vector<bool> avail(num.size(), true);
+        sort(num.begin(), num.end());
+        bool avail[num.size()];
+        memset(avail, true, sizeof(avail));
         vector<int> pum;
-        permuteUniqueRe(num, avail, pum);
+        permuteUniqueRe(num, pum, avail);
         return res;
     }
 
-    void permuteUniqueRe(vector<int> &num, vector<bool> &avail, vector<int> &pum)
+    void permuteUniqueRe(vector<int> &num, vector<int> &pum, bool avail[])
     {
         if (pum.size() == num.size())
         {
             res.push_back(pum);
             return;
         }
-        set<int> used;
+        int last_index = -1;
         for (int i = 0; i < num.size(); ++i)
         {
-            if (avail[i] && used.find(num[i]) == used.end())
-            {
-                avail[i] = false;
-                pum.push_back(num[i]);
-                permuteUniqueRe(num, avail, pum);
-                pum.pop_back();
-                avail[i] = true;
-                used.insert(num[i]);
-            }
+            if (!avail[i]) continue;
+            if (last_index != -1 && num[i] == num[last_index]) continue;
+
+            avail[i] = false;
+            pum.push_back(num[i]);
+            permuteUniqueRe(num, pum, avail);
+            pum.pop_back();
+            avail[i] = true;
+            last_index = i;
         }
     }
 };