From 391bc122543a74befce2cbcc8f1a86137e63071d Mon Sep 17 00:00:00 2001
From: YDZ <ydz627@gmail.com>
Date: Mon, 11 Mar 2019 00:18:54 +0800
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20problem=20143?=
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---
 .../143.Reorder List/143. Reorder List.go     | 66 +++++++++++++++++++
 .../143. Reorder List_test.go                 | 56 ++++++++++++++++
 Algorithms/143.Reorder List/README.md         | 30 +++++++++
 3 files changed, 152 insertions(+)
 create mode 100644 Algorithms/143.Reorder List/143. Reorder List.go
 create mode 100644 Algorithms/143.Reorder List/143. Reorder List_test.go
 create mode 100644 Algorithms/143.Reorder List/README.md

diff --git a/Algorithms/143.Reorder List/143. Reorder List.go b/Algorithms/143.Reorder List/143. Reorder List.go
new file mode 100644
index 00000000..2fae72a9
--- /dev/null
+++ b/Algorithms/143.Reorder List/143. Reorder List.go	
@@ -0,0 +1,66 @@
+package leetcode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func reorderList(head *ListNode) *ListNode {
+	if head == nil || head.Next == nil {
+		return head
+	}
+
+	// 寻找中间结点
+	p1 := head
+	p2 := head
+	for p2.Next != nil && p2.Next.Next != nil {
+		p1 = p1.Next
+		p2 = p2.Next.Next
+	}
+
+	// 反转链表后半部分  1->2->3->4->5->6 to 1->2->3->6->5->4
+	preMiddle := p1
+	preCurrent := p1.Next
+	for preCurrent.Next != nil {
+		current := preCurrent.Next
+		preCurrent.Next = current.Next
+		current.Next = preMiddle.Next
+		preMiddle.Next = current
+	}
+
+	// 重新拼接链表  1->2->3->6->5->4 to 1->6->2->5->3->4
+	p1 = head
+	p2 = preMiddle.Next
+	for p1 != preMiddle {
+		preMiddle.Next = p2.Next
+		p2.Next = p1.Next
+		p1.Next = p2
+		p1 = p2.Next
+		p2 = preMiddle.Next
+	}
+	return head
+}
+
+func reorderList_1(head *ListNode) *ListNode {
+	array := listToArray(head)
+	length := len(array)
+	if length == 0 {
+		return head
+	}
+	cur := head
+	last := head
+	for i := 0; i < len(array)/2; i++ {
+		tmp := &ListNode{Val: array[length-1-i], Next: cur.Next}
+		cur.Next = tmp
+		cur = tmp.Next
+		last = tmp
+	}
+	if length%2 == 0 {
+		last.Next = nil
+	} else {
+		cur.Next = nil
+	}
+	return head
+}
diff --git a/Algorithms/143.Reorder List/143. Reorder List_test.go b/Algorithms/143.Reorder List/143. Reorder List_test.go
new file mode 100644
index 00000000..d9a8fb9a
--- /dev/null
+++ b/Algorithms/143.Reorder List/143. Reorder List_test.go	
@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question143 struct {
+	para143
+	ans143
+}
+
+// para 是参数
+// one 代表第一个参数
+type para143 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans143 struct {
+	one []int
+}
+
+func Test_Problem143(t *testing.T) {
+
+	qs := []question143{
+
+		question143{
+			para143{[]int{1, 2, 3, 4, 5}},
+			ans143{[]int{1, 5, 2, 4, 3}},
+		},
+		question143{
+			para143{[]int{1, 2, 3, 4}},
+			ans143{[]int{1, 4, 2, 3}},
+		},
+
+		question143{
+			para143{[]int{1}},
+			ans143{[]int{1}},
+		},
+
+		question143{
+			para143{[]int{}},
+			ans143{[]int{}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 143------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans143, q.para143
+		fmt.Printf("【input】:%v       【output】:%v\n", p, L2s(reorderList(S2l(p.one))))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/143.Reorder List/README.md b/Algorithms/143.Reorder List/README.md
new file mode 100644
index 00000000..592b4f9b
--- /dev/null
+++ b/Algorithms/143.Reorder List/README.md	
@@ -0,0 +1,30 @@
+# [143. Reorder List](https://leetcode.com/problems/reorder-list/)
+
+## 题目
+
+Given a singly linked list L: L0→L1→…→Ln-1→Ln,
+reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
+
+You may not modify the values in the list's nodes, only nodes itself may be changed.
+
+Example 1:
+
+```c
+Given 1->2->3->4, reorder it to 1->4->2->3.
+```
+
+Example 2:
+
+```c
+Given 1->2->3->4->5, reorder it to 1->5->2->4->3.
+```
+
+## 题目大意
+
+按照指定规则重新排序链表：第一个元素和最后一个元素排列在一起，接着第二个元素和倒数第二个元素排在一起，接着第三个元素和倒数第三个元素排在一起。
+
+最近简单的方法是先把链表存储到数组里，然后找到链表中间的结点，按照规则拼接即可。这样时间复杂度是 O(n)，空间复杂度是 O(n)。
+
+更好的做法是结合之前几道题的操作：链表逆序，找中间结点。
+
+先找到链表的中间结点，然后利用逆序区间的操作，如 [92 题](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/92.Reverse%20Linked%20List%20II) 里的 reverseBetween() 操作，只不过这里的反转区间是从中点一直到末尾。最后利用 2 个指针，一个指向头结点，一个指向中间结点，开始拼接最终的结果。这种做法的时间复杂度是 O(n)，空间复杂度是 O(1)。
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