Add solution 143 [Java]

Author: yanglbme

README.md

@@ -69,6 +69,7 @@ Complete [solutions](https://github.com/doocs/leetcode/tree/master/solution) to
 | 130 | [Surrounded Regions](https://github.com/doocs/leetcode/tree/master/solution/130.Surrounded%20Regions) | `Depth-first Search`, `Breadth-first Search`, `Union Find` |
 | 137 | [Single Number II](https://github.com/doocs/leetcode/tree/master/solution/137.Single%20Number%20II) | `Bit Manipulation` |
 | 142 | [Linked List Cycle II](https://github.com/doocs/leetcode/tree/master/solution/142.Linked%20List%20Cycle%20II) | `Linked List`, `Two Pointers` |
+| 143| [Reorder List](https://github.com/doocs/leetcode/tree/master/solution/143.Reorder%20List) | `Linked List` |
 | 144 | [Binary Tree Preorder Traversal](https://github.com/doocs/leetcode/tree/master/solution/144.Binary%20Tree%20Preorder%20Traversal) | `Stack`, `Tree` |
 | 150 | [Evaluate Reverse Polish Notation](https://github.com/doocs/leetcode/tree/master/solution/150.Evaluate%20Reverse%20Polish%20Notation) | `Stack` |
 | 153 | [Find Minimum in Rotated Sorted Array](https://github.com/doocs/leetcode/tree/master/solution/153.Find%20Minimum%20in%20Rotated%20Sorted%20Array) | `Array`, `Binary Search` |

solution/143.Reorder List/README.md

@@ -0,0 +1,88 @@
+## 重排链表
+### 题目描述
+
+给定一个单链表 L：L0→L1→…→Ln-1→Ln ，
+将其重新排列后变为： L0→Ln→L1→Ln-1→L2→Ln-2→…
+
+你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。
+
+**示例 1:**
+```
+给定链表 1->2->3->4, 重新排列为 1->4->2->3.
+```
+
+**示例 2:**
+```
+给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.
+```
+
+### 解法
+先利用快慢指针找到链表的中间节点，之后对右半部分的链表进行`reverse`。最后对两个链表进行合并。注意边界的判断。
+
+
+```java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public void reorderList(ListNode head) {
+        if (head == null || head.next == null || head.next.next == null) {
+            return;
+        }
+        
+        ListNode dummy = new ListNode(-1);
+        dummy.next = head;
+        ListNode slow = head;
+        ListNode fast = head;
+        ListNode pre = dummy;
+        while (fast != null && fast.next != null) {
+            fast = fast.next.next;
+            slow = slow.next;
+            pre = pre.next;
+        }
+        
+        pre.next = null;
+        
+        // 将后半段的链表进行 reverse
+        ListNode rightPre = new ListNode(-1);
+        rightPre.next = null;
+        ListNode t = null;
+        while (slow != null) {
+            t = slow.next;
+            slow.next = rightPre.next;
+            rightPre.next = slow;
+            slow = t;
+        }
+        
+        ListNode p1 = dummy.next;
+        ListNode p2 = p1.next;
+        ListNode p3 = rightPre.next;
+        ListNode p4 = p3.next;
+        while (p1 != null) {
+            p3.next = p1.next;
+            p1.next = p3;
+            if (p2 == null) {
+                break;
+            }
+            p1 = p2;
+            p2 = p1.next;
+            p3 = p4;
+            p4 = p3.next;
+            
+        }
+        
+        if (p4 != null) {
+            p1.next.next = p4;
+        }
+        head = dummy.next;
+        
+        
+        
+    }
+}
+```

solution/143.Reorder List/Solution.java

@@ -0,0 +1,64 @@
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public void reorderList(ListNode head) {
+        if (head == null || head.next == null || head.next.next == null) {
+            return;
+        }
+        
+        ListNode dummy = new ListNode(-1);
+        dummy.next = head;
+        ListNode slow = head;
+        ListNode fast = head;
+        ListNode pre = dummy;
+        while (fast != null && fast.next != null) {
+            fast = fast.next.next;
+            slow = slow.next;
+            pre = pre.next;
+        }
+        
+        pre.next = null;
+        
+        // 将后半段的链表进行 reverse
+        ListNode rightPre = new ListNode(-1);
+        rightPre.next = null;
+        ListNode t = null;
+        while (slow != null) {
+            t = slow.next;
+            slow.next = rightPre.next;
+            rightPre.next = slow;
+            slow = t;
+        }
+        
+        ListNode p1 = dummy.next;
+        ListNode p2 = p1.next;
+        ListNode p3 = rightPre.next;
+        ListNode p4 = p3.next;
+        while (p1 != null) {
+            p3.next = p1.next;
+            p1.next = p3;
+            if (p2 == null) {
+                break;
+            }
+            p1 = p2;
+            p2 = p1.next;
+            p3 = p4;
+            p4 = p3.next;
+            
+        }
+        
+        if (p4 != null) {
+            p1.next.next = p4;
+        }
+        head = dummy.next;
+        
+        
+        
+    }
+}