From 8a24c8740c9b30ff5e9d263abe7e3e89f1e759a3 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <tangshusen@pku.edu.cn>
Date: Sat, 12 Jan 2019 12:29:18 +0800
Subject: [PATCH] Create 46. Permutations.md

---
 solutions/46. Permutations.md | 61 +++++++++++++++++++++++++++++++++++
 1 file changed, 61 insertions(+)
 create mode 100644 solutions/46. Permutations.md

diff --git a/solutions/46. Permutations.md b/solutions/46. Permutations.md
new file mode 100644
index 0000000..dbe09de
--- /dev/null
+++ b/solutions/46. Permutations.md	
@@ -0,0 +1,61 @@
+# [46. Permutations](https://leetcode.com/problems/permutations/)
+# 思路
+返回一个数组的所有排列，数组中元素没有重复。    
+如果做了[31. Next Permutation](https://leetcode.com/problems/next-permutation/)的话那这题就没什么问题了。稍有不同的是此题的数组元素没有重复而39题中可能重复。     
+所以此题可以使用STL中现成的next_permutation函数：
+> bool next_permutation (first, last)返回的是bool型，如果已经是最后一个排列了则返回false并将数组变成第一个排列(即按照从小到大排好序)；否则返回true并将数组变成下一个排列。
+此外还可以传入comp参数: next_permutation (first, last, comp)这样就可以自定义数组的大小规则。
+
+另外也可手动实现这个函数，参见我的[31题题解](https://github.com/ShusenTang/LeetCode/blob/master/solutions/31.%20Next%20Permutation.md)。   
+亲测使用STL中的要比手动实现慢很多。  
+
+# C++
+## 使用STL中的next_permutation
+``` C++
+class Solution {
+public:
+    vector<vector<int>> permute(vector<int>& nums) {
+        vector<vector<int>>res;
+        if(nums.empty()) return res;
+        sort(nums.begin(), nums.end()); // 先排序
+        res.push_back(nums);
+        while(next_permutation(nums.begin(), nums.end())) res.push_back(nums);
+        return res;
+    }
+};
+```
+## 手动实现(亲测更快)
+``` C++
+class Solution {
+private:
+    bool my_next_permute(vector<int>& nums){
+        int len = nums.size();
+        int i = len - 1;
+        while(i > 0 && nums[i] < nums[i - 1]) i--;
+        if(i == 0) return false;
+        
+        int mid, low = i, high = len - 1;
+        while(low <= high){ // 二分查找
+            mid = low + (high - low) / 2;
+            if(nums[mid] < nums[i - 1]) high = mid - 1;
+            else low = mid + 1;
+        }
+        
+        // int high = len - 1;
+        // while(nums[i - 1] > nums[high]) high--;
+        
+        swap(nums[i - 1], nums[high]);
+        reverse(nums.begin() + i, nums.end());
+        return true;
+    }
+public:
+    vector<vector<int>> permute(vector<int>& nums) {
+        vector<vector<int>>res;
+        if(nums.empty()) return res;
+        sort(nums.begin(), nums.end()); // 先排序
+        res.push_back(nums);
+        while(my_next_permute(nums)) res.push_back(nums);
+        return res;
+    }
+};
+```