Add solution 031

Author: yanglbme

README.md

@@ -41,12 +41,12 @@ Complete solutions to Leetcode problems, updated daily.
 | 015 | [3Sum](https://github.com/yanglbme/leetcode/tree/master/solution/015.3Sum) | `Array`, `Two Pointers` |
 | 019 | [Remove Nth Node From End of List](https://github.com/yanglbme/leetcode/tree/master/solution/019.Remove%20Nth%20Node%20From%20End%20of%20List) | `Linked List`, `Two Pointers` |
 | 024 | [Swap Nodes in Pairs](https://github.com/yanglbme/leetcode/tree/master/solution/024.Swap%20Nodes%20in%20Pairs) | `Linked List` |
+| 031 | [Next Permutation](https://github.com/yanglbme/leetcode/tree/master/solution/031.Next%20Permutation) | `Array` |
 | 046 | [Permutations](https://github.com/yanglbme/leetcode/tree/master/solution/046.Permutations) | `Backtracking` |
 | 047 | [Permutations II](https://github.com/yanglbme/leetcode/tree/master/solution/047.Permutations%20II) | `Backtracking` |
 | 062 | [Unique Paths](https://github.com/yanglbme/leetcode/tree/master/solution/062.Unique%20Paths) | `Array`, `Dynamic Programming` |
 | 063 | [Unique Paths II](https://github.com/yanglbme/leetcode/tree/master/solution/063.Unique%20Paths%20II) | `Array`, `Dynamic Programming` |
 | 075 | [Sort Colors](https://github.com/yanglbme/leetcode/tree/master/solution/075.Sort%20Colors) | `Array`, `Two Pointers`, `Sort` |
-075.Sort Colors
 | 082 | [Remove Duplicates from Sorted List II](https://github.com/yanglbme/leetcode/tree/master/solution/082.Remove%20Duplicates%20from%20Sorted%20List%20II) | `Linked List` |
 | 153 | [Find Minimum in Rotated Sorted Array](https://github.com/yanglbme/leetcode/tree/master/solution/153.Find%20Minimum%20in%20Rotated%20Sorted%20Array) | `Array`, `Binary Search` |
 

solution/031.Next Permutation/README.md

@@ -0,0 +1,72 @@
+## 下一个排列
+### 题目描述
+
+实现获取下一个排列的函数，算法需要将给定数字序列重新排列成字典序中下一个更大的排列。
+
+如果不存在下一个更大的排列，则将数字重新排列成最小的排列（即升序排列）。
+
+必须原地修改，只允许使用额外常数空间。
+
+以下是一些例子，输入位于左侧列，其相应输出位于右侧列。
+
+`1,2,3` → `1,3,2`
+
+`3,2,1` → `1,2,3`
+
+`1,1,5` → `1,5,1`
+
+### 解法
+从后往前，找到第一个升序状态的位置，记为 i-1，在[i, length - 1] 中找到比 nums[i - 1] 大的，且差值最小的元素（如果有多个差值最小且相同的元素，取后者），进行交换。将后面的数组序列升序排列，保存。然后恢复 i，继续循环。
+
+
+```java
+class Solution {
+    public void nextPermutation(int[] nums) {
+        boolean flag = false;
+        for (int i = nums.length - 2; i >= 0; --i) {
+            if (nums[i] < nums[i + 1]) {
+                int index = findMinIndex(nums, i, nums[i]);
+                swap(nums, i, index);
+                reverse(nums, i + 1);
+                flag = true;
+                break;
+            }
+        }
+        if (!flag) {
+            Arrays.sort(nums);
+        }
+    }
+    
+    private void reverse(int[] nums, int start) {
+        int end = nums.length - 1;
+        while (start < end) {
+            swap(nums, start++, end--);
+        }
+    }
+    
+    /**
+     * 找出从start开始的比val大的最小元素的下标，如果有多个，选择后者 
+     *
+     * @param name
+     * @param start
+     * @param val
+     * @return index
+     */
+    private int findMinIndex(int[] nums, int start, int val) {
+        int end = nums.length - 1;
+        int i = start;
+        for (; i < end; ++i) {
+            if (nums[i + 1] <= val) {
+                break;
+            }
+        }
+        return i;
+    }
+    
+    private void swap(int[] nums, int i, int j) {
+        int t = nums[i];
+        nums[i] = nums[j];
+        nums[j] = t;
+    }
+}
+```

solution/031.Next Permutation/Solution.java

@@ -0,0 +1,49 @@
+class Solution {
+    public void nextPermutation(int[] nums) {
+        boolean flag = false;
+        for (int i = nums.length - 2; i >= 0; --i) {
+            if (nums[i] < nums[i + 1]) {
+                int index = findMinIndex(nums, i, nums[i]);
+                swap(nums, i, index);
+                reverse(nums, i + 1);
+                flag = true;
+                break;
+            }
+        }
+        if (!flag) {
+            Arrays.sort(nums);
+        }
+    }
+    
+    private void reverse(int[] nums, int start) {
+        int end = nums.length - 1;
+        while (start < end) {
+            swap(nums, start++, end--);
+        }
+    }
+    
+    /**
+     * 找出从start开始的比val大的最小元素的下标，如果有多个，选择后者 
+     *
+     * @param name
+     * @param start
+     * @param val
+     * @return index
+     */
+    private int findMinIndex(int[] nums, int start, int val) {
+        int end = nums.length - 1;
+        int i = start;
+        for (; i < end; ++i) {
+            if (nums[i + 1] <= val) {
+                break;
+            }
+        }
+        return i;
+    }
+    
+    private void swap(int[] nums, int i, int j) {
+        int t = nums[i];
+        nums[i] = nums[j];
+        nums[j] = t;
+    }
+}