Add solution 086

Author: yanglbme

README.md

@@ -50,6 +50,7 @@ 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` |

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 链表种，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

@@ -0,0 +1,25 @@
+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;
+
+    }
+}