feat(leetcode): 206

Codes/StarDrewer/206_reverse-linked-list/solve.go

@@ -0,0 +1,101 @@
+//Given the head of a singly linked list, reverse the list, and return the
+//reversed list.
+//
+//
+// Example 1:
+//
+//
+//Input: head = [1,2,3,4,5]
+//Output: [5,4,3,2,1]
+//
+//
+// Example 2:
+//
+//
+//Input: head = [1,2]
+//Output: [2,1]
+//
+//
+// Example 3:
+//
+//
+//Input: head = []
+//Output: []
+//
+//
+//
+// Constraints:
+//
+//
+// The number of nodes in the list is the range [0, 5000].
+// -5000 <= Node.val <= 5000
+//
+//
+//
+// Follow up: A linked list can be reversed either iteratively or recursively.
+//Could you implement both?
+// Related Topics 递归 链表 👍 1920 👎 0
+
+package main
+
+import "fmt"
+
+type ListNode struct {
+	Val  int
+	Next *ListNode
+}
+
+//leetcode submit region begin(Prohibit modification and deletion)
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+
+// 为了方便阅读，单独写了一个链表节点索引结构
+type IndexNode struct {
+	Node *ListNode
+}
+
+func reverseList(head *ListNode) *ListNode {
+	if head == nil {
+		return nil
+	}
+
+	var l, r IndexNode
+	l.Node = head
+	r.Node = head.Next
+
+	l.Node.Next = nil // 使头结点变尾结点
+
+	for r.Node != nil {
+		tmp := r.Node.Next   // 保存右链index的下一位
+		r.Node.Next = l.Node // 使右链index的节点指向左链
+		l.Node = r.Node      // 左链index++
+		r.Node = tmp         // 右链index++
+	}
+
+	return l.Node
+}
+
+//leetcode submit region end(Prohibit modification and deletion)
+
+func main() {
+	res := reverseList(&ListNode{
+		Val: 0,
+		Next: &ListNode{
+			Val: 1,
+			Next: &ListNode{
+				Val:  2,
+				Next: nil,
+			},
+		},
+	})
+	for p := res; p != nil; p = p.Next {
+		fmt.Printf("->%v", p.Val)
+	}
+	fmt.Printf("\n")
+
+}