another solution

Permutations.h

@@ -12,6 +12,10 @@
  [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].
 
  Solution: dfs...
+ Solution2：non-recursive and unique result. Find the next permutation and add it to result set, 
+			this will miss permutations that 'smaller' than the given 'nums'. 
+			So, at first,we sort the nums. This is the fastest one in all test case. the dfs's
+			result may contain duplicate permutations.
  */
 
 class Solution {
@@ -46,3 +50,35 @@ class Solution {
         }
     }
 };
+
+class Solution2 {
+public:
+	//return all possible unique permutations.
+	vector<vector<int>> permute(vector<int>& nums) {
+		sort(nums.begin(), nums.end());//!!!
+		vector<vector<int>> ret;
+		ret.push_back(nums);
+		while (1)
+		{
+			int idx = -1;
+			for (int i = nums.size() - 1; i > 0; i--){
+				if (nums[i] > nums[i - 1]){
+					idx = i - 1;
+					break;
+				}
+			}
+			if (idx == -1) break;
+			for (int i = nums.size() - 1; i > 0; i--){
+				if (nums[i] > nums[idx]){
+					int tmp = nums[i];
+					nums[i] = nums[idx];
+					nums[idx] = tmp;
+					break;
+				}
+			}
+			sort(&nums[0] + idx + 1, &nums[0] + nums.size());
+			ret.push_back(nums);
+		}
+		return ret;
+	}
+};

PermutationsII.h

@@ -12,6 +12,9 @@
  [1,1,2], [1,2,1], and [2,1,1].
 
  Solution: dfs...
+ Solution2：non-recursive. Find the next permutation and add it to result set, 
+			this will miss permutations that 'smaller' than the given 'nums'. 
+			So, at first,we sort the nums. This is the fastest one in all test case. 
  */
 
 class Solution {
@@ -49,3 +52,35 @@ class Solution {
         }
     }
 };
+
+class Solution2 {
+public:
+	//return all possible unique permutations.
+	vector<vector<int>> permuteUnique(vector<int>& nums) {
+		sort(nums.begin(), nums.end());//!!!
+		vector<vector<int>> ret;
+		ret.push_back(nums);
+		while (1)
+		{
+			int idx = -1;
+			for (int i = nums.size() - 1; i > 0; i--){
+				if (nums[i] > nums[i - 1]){
+					idx = i - 1;
+					break;
+				}
+			}
+			if (idx == -1) break;
+			for (int i = nums.size() - 1; i > 0; i--){
+				if (nums[i] > nums[idx]){
+					int tmp = nums[i];
+					nums[i] = nums[idx];
+					nums[idx] = tmp;
+					break;
+				}
+			}
+			sort(&nums[0] + idx + 1, &nums[0] + nums.size());
+			ret.push_back(nums);
+		}
+		return ret;
+	}
+};