add Next Permutation

Author: billryan

SUMMARY.md

@@ -111,6 +111,7 @@
    * [Permutation](exhaustive_search/permutation.md)
    * [Unique Permutations](exhaustive_search/unique_permutations.md)
    * [Unique Binary Search Trees II](exhaustive_search/unique_binary_search_trees_ii.md)
+   * [Next Permutation](exhaustive_search/next_permutation.md)
 * [Dynamic Programming - 动态规划](dynamic_programming/README.md)
    * [Triangle](dynamic_programming/triangle.md)
    * [Knapsack - 背包问题](dynamic_programming/knapsack.md)

exhaustive_search/next_permutation.md

@@ -0,0 +1,174 @@
+# Next Permutation
+
+## Source
+
+- lintcode: [(52) Next Permutation](http://www.lintcode.com/en/problem/next-permutation/)
+
+```
+Given a list of integers, which denote a permutation.
+
+Find the next permutation in ascending order.
+
+Example
+For [1,3,2,3], the next permutation is [1,3,3,2]
+
+For [4,3,2,1], the next permutation is [1,2,3,4]
+
+Note
+The list may contains duplicate integers.
+```
+
+## 题解
+
+找下一个升序排列，C++ STL 源码剖析一书中有提及，[Permutation](http://algorithm.yuanbin.me/exhaustive_search/permutation.html) 一小节中也有详细介绍，下面简要介绍一下字典序算法：
+
+1. 从后往前寻找索引满足 `a[k] < a[k + 1]`, 如果此条件不满足，则说明已遍历到最后一个。
+2. 从后往前遍历，找到第一个比`a[k]`大的数`a[l]`, 即`a[k] < a[l]`.
+3. 交换`a[k]`与`a[l]`.
+4. 反转`k + 1 ~ n`之间的元素。
+
+由于这道题中规定对于`[4,3,2,1]`, 输出为`[1,2,3,4]`, 故在第一步稍加处理即可。
+
+### Python
+
+```python
+class Solution:
+    # @param num :  a list of integer
+    # @return : a list of integer
+    def nextPermutation(self, num):
+        if num is None or len(num) <= 1:
+            return num
+        # step1: find nums[i] < nums[i + 1], Loop backwards
+        i = 0
+        for i in xrange(len(num) - 2, -1, -1):
+            if num[i] < num[i + 1]:
+                break
+            elif i == 0:
+                # reverse nums if reach maximum
+                num = num[::-1]
+                return num
+        # step2: find nums[i] < nums[j], Loop backwards
+        j = 0
+        for j in xrange(len(num) - 1, i, -1):
+            if num[i] < num[j]:
+                break
+        # step3: swap betwenn nums[i] and nums[j]
+        num[i], num[j] = num[j], num[i]
+        # step4: reverse between [i + 1, n - 1]
+        num[i + 1:len(num)] = num[len(num) - 1:i:-1]
+
+        return num
+```
+
+### C++
+
+```c++
+class Solution {
+public:
+    /**
+     * @param nums: An array of integers
+     * @return: An array of integers that's next permuation
+     */
+    vector<int> nextPermutation(vector<int> &nums) {
+        if (nums.empty() || nums.size() <= 1) {
+            return nums;
+        }
+        // step1: find nums[i] < nums[i + 1]
+        int i = 0;
+        for (i = nums.size() - 2; i >= 0; --i) {
+            if (nums[i] < nums[i + 1]) {
+                break;
+            } else if (0 == i) {
+                // reverse nums if reach maximum
+                reverse(nums, 0, nums.size() - 1);
+                return nums;
+            }
+        }
+        // step2: find nums[i] < nums[j]
+        int j = 0;
+        for (j = nums.size() - 1; j > i; --j) {
+            if (nums[i] < nums[j]) break;
+        }
+        // step3: swap betwenn nums[i] and nums[j]
+        int temp = nums[i];
+        nums[i] = nums[j];
+        nums[j] = temp;
+        // step4: reverse between [i + 1, n - 1]
+        reverse(nums, i + 1, nums.size() - 1);
+
+        return nums;
+
+    }
+
+private:
+    void reverse(vector<int>& nums, int start, int end) {
+        for (int i = start, j = end; i < j; ++i, --j) {
+            int temp = nums[i];
+            nums[i] = nums[j];
+            nums[j] = temp;
+        }
+    }
+};
+```
+
+### Java
+
+```java
+public class Solution {
+    /**
+     * @param nums: an array of integers
+     * @return: return nums in-place
+     */
+    public int[] nextPermutation(int[] nums) {
+        if (nums == null || nums.length <= 1) {
+            return nums;
+        }
+        // step1: find nums[i] < nums[i + 1]
+        int i = 0;
+        for (i = nums.length - 2; i >= 0; i--) {
+            if (nums[i] < nums[i + 1]) {
+                break;
+            } else if (i == 0) {
+                // reverse nums if reach maximum
+                reverse(nums, 0, nums.length - 1);
+                return nums;
+            }
+        }
+        // step2: find nums[i] < nums[j]
+        int j = 0;
+        for (j = nums.length - 1; j > i; j--) {
+            if (nums[i] < nums[j]) {
+                break;
+            }
+        }
+        // step3: swap betwenn nums[i] and nums[j]
+        int temp = nums[i];
+        nums[i] = nums[j];
+        nums[j] = temp;
+        // step4: reverse between [i + 1, n - 1]
+        reverse(nums, i + 1, nums.length - 1);
+
+        return nums;
+    }
+
+    private void reverse(int[] nums, int start, int end) {
+        for (int i = start, j = end; i < j; i++, j--) {
+            int temp = nums[i];
+            nums[i] = nums[j];
+            nums[j] = temp;
+        }
+    }
+}
+```
+
+### 源码分析
+
+和 Permutation 一小节类似，这里只需要注意在step 1中`i == 0`时需要反转之以获得最小的序列。对于有重复元素，只要在 step1和 step2中判断元素大小时不取等号即可。
+
+### 复杂度分析
+
+最坏情况下，遍历两次原数组，反转一次数组，时间复杂度为 $$O(n)$$, 使用了 temp 临时变量，空间复杂度可认为是 $$O(1)$$.
+
+## Reference
+
+- [Permutation | Data Structure and Algorithm](http://algorithm.yuanbin.me/exhaustive_search/permutation.html)