diff --git a/README.md b/README.md
index 1eab29d..32aea76 100644
--- a/README.md
+++ b/README.md
@@ -6,7 +6,9 @@
 + [7 Reverse Integer(溢出检测)](algorithms/ReverseInteger)
 + [8 String to Integer (atoi)(溢出检测）](algorithms/StringtoInteger)
 + [21 Merge Two Sorted Lists](algorithms/MergeTwoSortedLists)
-+ [31 Next Permutation](algorithms/NextPermutation)
++ [31 Next Permutation(全排列)](algorithms/NextPermutation)
++ [46 Permutations(全排列)](algorithms/Permutations)
++ [47 Permutations II(全排列)](algorithms/Permutations2)
 + [60 Permutation Sequence(康托展开)](algorithms/PermutationSequence)
 + [65 Valid Number(细节题,状态机)](algorithms/ValidNumber)
 + [191 Number of 1 Bits(位运算)](algorithms/Numberof1Bits)
diff --git a/algorithms/Permutations/README.md b/algorithms/Permutations/README.md
new file mode 100644
index 0000000..48992ff
--- /dev/null
+++ b/algorithms/Permutations/README.md
@@ -0,0 +1,43 @@
+## Permutations
+
+Given a collection of numbers, return all possible permutations.
+
+For example,
+[1,2,3] have the following permutations:
+[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].
+
+## Solution
+
+全排列, [从一个排列求下一个排列算法](https://github.com/krystism/algorithms/tree/master/permutation)
+
+## Code
+```cpp
+class Solution {
+	public:
+		    vector<vector<int> > permute(vector<int> &num) {
+				vector<vector<int>> ans;
+				sort(num.begin(), num.end());
+				do {
+					vector<int> v(num.begin(), num.end());
+					ans.push_back(v);
+				} while(next(num));
+				return ans;
+			}
+	private:
+		   	bool next(vector<int> &v) {
+				size_t n = v.size();
+				int i;
+				for (i = n - 2; i >= 0 && v[i] >= v[i + 1]; --i);
+				if (i < 0) {
+					reverse(v.begin(), v.end());
+					return false;
+				}
+				int j;
+				for (j = n - 1; j > i && v[j] <= v[i]; --j);
+				swap(v[i], v[j]);
+				reverse(v.begin() + i + 1, v.end());
+				return true;
+			}
+
+};
+```
diff --git a/algorithms/Permutations/permutations.cpp b/algorithms/Permutations/permutations.cpp
new file mode 100644
index 0000000..0d4fbda
--- /dev/null
+++ b/algorithms/Permutations/permutations.cpp
@@ -0,0 +1,48 @@
+#include <vector>
+#include <algorithm>
+#include <cstdio>
+#include <cstdlib>
+using namespace std;
+void print(vector<int> v)
+{
+	for (auto i : v)
+		printf("%d ", i);
+	printf("\n");
+}
+class Solution {
+	public:
+		    vector<vector<int> > permute(vector<int> &num) {
+				vector<vector<int>> ans;
+				sort(num.begin(), num.end());
+				do {
+					vector<int> v(num.begin(), num.end());
+					ans.push_back(v);
+				} while(next(num));
+				return ans;
+			}
+	private:
+		   	bool next(vector<int> &v) {
+				size_t n = v.size();
+				int i;
+				for (i = n - 2; i >= 0 && v[i] >= v[i + 1]; --i);
+				if (i < 0) {
+					reverse(v.begin(), v.end());
+					return false;
+				}
+				int j;
+				for (j = n - 1; j > i && v[j] <= v[i]; --j);
+				swap(v[i], v[j]);
+				reverse(v.begin() + i + 1, v.end());
+				return true;
+			}
+
+};
+int main(int argc, char **argv)
+{
+	Solution solution;
+	vector<int> num = {1,2,3, 4, 5};
+	vector<vector<int>> ans = solution.permute(num);
+	for (auto i : ans)
+		print(i);
+	return 0;
+}
diff --git a/algorithms/Permutations2/README.md b/algorithms/Permutations2/README.md
new file mode 100644
index 0000000..493a3da
--- /dev/null
+++ b/algorithms/Permutations2/README.md
@@ -0,0 +1,43 @@
+## 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]`.
+
+## Solution
+
+全排列, [从一个排列求下一个排列算法](https://github.com/krystism/algorithms/tree/master/permutation)
+
+## Code
+```cpp
+class Solution {
+	public:
+		    vector<vector<int> > permuteUnique(vector<int> &num) {
+				vector<vector<int>> ans;
+				sort(num.begin(), num.end());
+				do {
+					vector<int> v(num.begin(), num.end());
+					ans.push_back(v);
+				} while(next(num));
+				return ans;
+			}
+	private:
+		   	bool next(vector<int> &v) {
+				size_t n = v.size();
+				int i;
+				for (i = n - 2; i >= 0 && v[i] >= v[i + 1]; --i);
+				if (i < 0) {
+					reverse(v.begin(), v.end());
+					return false;
+				}
+				int j;
+				for (j = n - 1; j > i && v[j] <= v[i]; --j);
+				swap(v[i], v[j]);
+				reverse(v.begin() + i + 1, v.end());
+				return true;
+			}
+
+};
+```