Merge branch 'master' of github.com:yanglbme/leetcode

Author: chakyam

README.md

@@ -50,6 +50,8 @@ Complete solutions to Leetcode problems, updated daily.
 | 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` |
 | 082 | [Remove Duplicates from Sorted List II](https://github.com/yanglbme/leetcode/tree/master/solution/082.Remove%20Duplicates%20from%20Sorted%20List%20II) | `Linked List` |
+| 086 | [Partition List](https://github.com/yanglbme/leetcode/tree/master/solution/086.Partition%20List) | `Linked List`, `Two Pointers` |
+| 092 | [Reverse Linked List II](https://github.com/yanglbme/leetcode/tree/master/solution/092.Reverse%20Linked%20List%20II) | `Linked List` |
 | 094 | [Binary Tree Inorder Traversal](https://github.com/yanglbme/leetcode/tree/master/solution/094.Binary%20Tree%20Inorder%20Traversal) | `Hash Table`, `Stack`, `Tree` |
 | 144 | [Binary Tree Preorder Traversal](https://github.com/yanglbme/leetcode/tree/master/solution/144.Binary%20Tree%20Preorder%20Traversal) | `Stack`, `Tree` |
 | 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/086.Partition List/README.md

@@ -0,0 +1,52 @@
+## 分隔链表
+### 题目描述
+
+给定一个链表和一个特定值 x，对链表进行分隔，使得所有小于 x 的节点都在大于或等于 x 的节点之前。
+
+你应当保留两个分区中每个节点的初始相对位置。
+
+示例:
+
+```
+输入: head = 1->4->3->2->5->2, x = 3
+输出: 1->2->2->4->3->5
+```
+
+### 解法
+维护 `left`, `right` 两个链表，遍历 `head` 链表，若对应元素值小于 `x`，将该结点插入 `left` 链表中，否则插入 `right` 链表中。最后 `right` 尾部指向空，并将 `left` 尾部指向 `right` 头部，使得它们串在一起。
+
+```java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode partition(ListNode head, int x) {
+        ListNode leftDummy = new ListNode(-1);
+        ListNode rightDummy = new ListNode(-1);
+        
+        ListNode leftCur = leftDummy;
+        ListNode rightCur = rightDummy;
+        
+        while (head != null) {
+            if (head.val < x) {
+                leftCur.next = head;
+                leftCur = leftCur.next;
+            } else {
+                rightCur.next = head;
+                rightCur = rightCur.next;
+            }
+            head = head.next;
+        }
+        
+        leftCur.next = rightDummy.next;
+        rightCur.next = null;
+        return leftDummy.next;
+        
+    }
+}
+```

solution/086.Partition List/Solution.java

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+class Solution {
+    public ListNode partition(ListNode head, int x) {
+        ListNode leftDummy = new ListNode(-1);
+        ListNode rightDummy = new ListNode(-1);
+
+        ListNode leftCur = leftDummy;
+        ListNode rightCur = rightDummy;
+
+        while (head != null) {
+            if (head.val < x) {
+                leftCur.next = head;
+                leftCur = leftCur.next;
+            } else {
+                rightCur.next = head;
+                rightCur = rightCur.next;
+            }
+            head = head.next;
+        }
+
+        leftCur.next = rightDummy.next;
+        rightCur.next = null;
+        return leftDummy.next;
+
+    }
+}

solution/092.Reverse Linked List II/README.md

@@ -0,0 +1,53 @@
+## 反转链表 II
+### 题目描述
+
+反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。
+
+说明:
+1 ≤ m ≤ n ≤ 链表长度。
+
+示例:
+
+```
+输入: 1->2->3->4->5->NULL, m = 2, n = 4
+输出: 1->4->3->2->5->NULL
+```
+
+### 解法
+利用头插法，对 [m + 1, n] 范围内的元素逐一插入。
+
+```java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode reverseBetween(ListNode head, int m, int n) {
+        ListNode dummy = new ListNode(-1);
+        dummy.next = head;
+        ListNode pre = dummy;
+        int i = 0;
+        for (; i < m - 1; ++i) {
+            pre = pre.next;
+        }
+        
+        ListNode head2 = pre;
+        pre = pre.next;
+        ListNode cur = pre.next;
+        for (; i < n - 1; ++i) {
+            pre.next = cur.next;
+            cur.next = head2.next;
+            head2.next = cur;
+            cur = pre.next;
+        }
+        
+        return dummy.next;
+        
+        
+    }
+}
+```

solution/092.Reverse Linked List II/Solution.java

@@ -0,0 +1,25 @@
+class Solution {
+    public ListNode reverseBetween(ListNode head, int m, int n) {
+        ListNode dummy = new ListNode(-1);
+        dummy.next = head;
+        ListNode pre = dummy;
+        int i = 0;
+        for (; i < m - 1; ++i) {
+            pre = pre.next;
+        }
+        
+        ListNode head2 = pre;
+        pre = pre.next;
+        ListNode cur = pre.next;
+        for (; i < n - 1; ++i) {
+            pre.next = cur.next;
+            cur.next = head2.next;
+            head2.next = cur;
+            cur = pre.next;
+        }
+        
+        return dummy.next;
+        
+        
+    }
+}