feat: update solutions to lc problem: No.0143 (#2547)

No.0143.Reorder List

Author: yanglbme

solution/0100-0199/0143.Reorder List/README.md

@@ -245,20 +245,28 @@ func reorderList(head *ListNode) {
  Do not return anything, modify head in-place instead.
  */
 function reorderList(head: ListNode | null): void {
-    const arr = [];
-    let node = head;
-    while (node.next != null) {
-        arr.push(node);
-        node = node.next;
+    let slow = head;
+    let fast = head;
+    // 找到中心节点
+    while (fast && fast.next) {
+        slow = slow.next;
+        fast = fast.next.next;
+    }
+    // 反转节点
+    let next = slow.next;
+    slow.next = null;
+    while (next) {
+        [next.next, slow, next] = [slow, next, next.next];
     }
-    let l = 0;
-    let r = arr.length - 1;
-    while (l < r) {
-        const start = arr[l];
-        const end = arr[r];
-        [end.next.next, start.next, end.next] = [start.next, end.next, null];
-        l++;
-        r--;
+    // 合并
+    let left = head;
+    let right = slow;
+    while (right.next) {
+        const next = left.next;
+        left.next = right;
+        right = right.next;
+        left.next.next = next;
+        left = left.next.next;
     }
 }
 ```
@@ -399,53 +407,4 @@ public class Solution {
 
 <!-- tabs:end -->
 
-### 方法二
-
-<!-- tabs:start -->
-
-```ts
-/**
- * Definition for singly-linked list.
- * class ListNode {
- *     val: number
- *     next: ListNode | null
- *     constructor(val?: number, next?: ListNode | null) {
- *         this.val = (val===undefined ? 0 : val)
- *         this.next = (next===undefined ? null : next)
- *     }
- * }
- */
-
-/**
- Do not return anything, modify head in-place instead.
- */
-function reorderList(head: ListNode | null): void {
-    let slow = head;
-    let fast = head;
-    // 找到中心节点
-    while (fast != null && fast.next != null) {
-        slow = slow.next;
-        fast = fast.next.next;
-    }
-    // 反转节点
-    let next = slow.next;
-    slow.next = null;
-    while (next != null) {
-        [next.next, slow, next] = [slow, next, next.next];
-    }
-    // 合并
-    let left = head;
-    let right = slow;
-    while (right.next != null) {
-        const next = left.next;
-        left.next = right;
-        right = right.next;
-        left.next.next = next;
-        left = left.next.next;
-    }
-}
-```
-
-<!-- tabs:end -->
-
 <!-- end -->

solution/0100-0199/0143.Reorder List/README_EN.md

@@ -45,7 +45,11 @@ L<sub>0</sub> &rarr; L<sub>n</sub> &rarr; L<sub>1</sub> &rarr; L<sub>n - 1</sub>
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Fast and Slow Pointers + Reverse List + Merge Lists
+
+We first use fast and slow pointers to find the midpoint of the linked list, then reverse the second half of the list, and finally merge the two halves.
+
+The time complexity is $O(n)$, where $n$ is the length of the linked list. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -57,26 +61,21 @@ L<sub>0</sub> &rarr; L<sub>n</sub> &rarr; L<sub>1</sub> &rarr; L<sub>n - 1</sub>
 #         self.next = next
 class Solution:
     def reorderList(self, head: Optional[ListNode]) -> None:
-        # 快慢指针找到链表中点
         fast = slow = head
         while fast.next and fast.next.next:
             slow = slow.next
             fast = fast.next.next
 
-        # cur 指向右半部分链表
         cur = slow.next
         slow.next = None
 
-        # 反转右半部分链表
         pre = None
         while cur:
             t = cur.next
             cur.next = pre
             pre, cur = cur, t
         cur = head
 
-        # 此时 cur, pre 分别指向链表左右两半的第一个节点
-        # 合并
         while pre:
             t = pre.next
             pre.next = cur.next
@@ -97,18 +96,15 @@ class Solution:
  */
 class Solution {
     public void reorderList(ListNode head) {
-        // 快慢指针找到链表中点
         ListNode fast = head, slow = head;
         while (fast.next != null && fast.next.next != null) {
             slow = slow.next;
             fast = fast.next.next;
         }
 
-        // cur 指向右半部分链表
         ListNode cur = slow.next;
         slow.next = null;
 
-        // 反转右半部分链表
         ListNode pre = null;
         while (cur != null) {
             ListNode t = cur.next;
@@ -118,8 +114,6 @@ class Solution {
         }
         cur = head;
 
-        // 此时 cur, pre 分别指向链表左右两半的第一个节点
-        // 合并
         while (pre != null) {
             ListNode t = pre.next;
             pre.next = cur.next;
@@ -145,19 +139,16 @@ class Solution {
 class Solution {
 public:
     void reorderList(ListNode* head) {
-        // 快慢指针找到链表中点
         ListNode* fast = head;
         ListNode* slow = head;
         while (fast->next && fast->next->next) {
             slow = slow->next;
             fast = fast->next->next;
         }
 
-        // cur 指向右半部分链表
         ListNode* cur = slow->next;
         slow->next = nullptr;
 
-        // 反转右半部分链表
         ListNode* pre = nullptr;
         while (cur) {
             ListNode* t = cur->next;
@@ -167,8 +158,6 @@ public:
         }
         cur = head;
 
-        // 此时 cur, pre 分别指向链表左右两半的第一个节点
-        // 合并
         while (pre) {
             ListNode* t = pre->next;
             pre->next = cur->next;
@@ -189,17 +178,14 @@ public:
  * }
  */
 func reorderList(head *ListNode) {
-	// 快慢指针找到链表中点
 	fast, slow := head, head
 	for fast.Next != nil && fast.Next.Next != nil {
 		slow, fast = slow.Next, fast.Next.Next
 	}
 
-	// cur 指向右半部分链表
 	cur := slow.Next
 	slow.Next = nil
 
-	// 反转右半部分链表
 	var pre *ListNode
 	for cur != nil {
 		t := cur.Next
@@ -208,8 +194,6 @@ func reorderList(head *ListNode) {
 	}
 	cur = head
 
-	// 此时 cur, pre 分别指向链表左右两半的第一个节点
-	// 合并
 	for pre != nil {
 		t := pre.Next
 		pre.Next = cur.Next
@@ -236,20 +220,27 @@ func reorderList(head *ListNode) {
  Do not return anything, modify head in-place instead.
  */
 function reorderList(head: ListNode | null): void {
-    const arr = [];
-    let node = head;
-    while (node.next != null) {
-        arr.push(node);
-        node = node.next;
+    let slow = head;
+    let fast = head;
+    while (fast && fast.next) {
+        slow = slow.next;
+        fast = fast.next.next;
     }
-    let l = 0;
-    let r = arr.length - 1;
-    while (l < r) {
-        const start = arr[l];
-        const end = arr[r];
-        [end.next.next, start.next, end.next] = [start.next, end.next, null];
-        l++;
-        r--;
+
+    let next = slow.next;
+    slow.next = null;
+    while (next) {
+        [next.next, slow, next] = [slow, next, next.next];
+    }
+
+    let left = head;
+    let right = slow;
+    while (right.next) {
+        const next = left.next;
+        left.next = right;
+        right = right.next;
+        left.next.next = next;
+        left = left.next.next;
     }
 }
 ```
@@ -305,19 +296,16 @@ impl Solution {
  * @return {void} Do not return anything, modify head in-place instead.
  */
 var reorderList = function (head) {
-    // 快慢指针找到链表中点
     let slow = head;
     let fast = head;
     while (fast.next && fast.next.next) {
         slow = slow.next;
         fast = fast.next.next;
     }
 
-    // cur 指向右半部分链表
     let cur = slow.next;
     slow.next = null;
 
-    // 反转右半部分链表
     let pre = null;
     while (cur) {
         const t = cur.next;
@@ -327,8 +315,6 @@ var reorderList = function (head) {
     }
     cur = head;
 
-    // 此时 cur, pre 分别指向链表左右两半的第一个节点
-    // 合并
     while (pre) {
         const t = pre.next;
         pre.next = cur.next;
@@ -353,19 +339,16 @@ var reorderList = function (head) {
  */
 public class Solution {
     public void ReorderList(ListNode head) {
-        // 快慢指针找到链表中点
         ListNode slow = head;
         ListNode fast = head;
         while (fast.next != null && fast.next.next != null) {
             slow = slow.next;
             fast = fast.next.next;
         }
 
-        // cur 指向右半部分链表
         ListNode cur = slow.next;
         slow.next = null;
 
-        // 反转右半部分链表
         ListNode pre = null;
         while (cur != null) {
             ListNode t = cur.next;
@@ -375,8 +358,6 @@ public class Solution {
         }
         cur = head;
 
-        // 此时 cur, pre 分别指向链表左右两半的第一个节点
-        // 合并
         while (pre != null) {
             ListNode t = pre.next;
             pre.next = cur.next;
@@ -390,53 +371,4 @@ public class Solution {
 
 <!-- tabs:end -->
 
-### Solution 2
-
-<!-- tabs:start -->
-
-```ts
-/**
- * Definition for singly-linked list.
- * class ListNode {
- *     val: number
- *     next: ListNode | null
- *     constructor(val?: number, next?: ListNode | null) {
- *         this.val = (val===undefined ? 0 : val)
- *         this.next = (next===undefined ? null : next)
- *     }
- * }
- */
-
-/**
- Do not return anything, modify head in-place instead.
- */
-function reorderList(head: ListNode | null): void {
-    let slow = head;
-    let fast = head;
-    // 找到中心节点
-    while (fast != null && fast.next != null) {
-        slow = slow.next;
-        fast = fast.next.next;
-    }
-    // 反转节点
-    let next = slow.next;
-    slow.next = null;
-    while (next != null) {
-        [next.next, slow, next] = [slow, next, next.next];
-    }
-    // 合并
-    let left = head;
-    let right = slow;
-    while (right.next != null) {
-        const next = left.next;
-        left.next = right;
-        right = right.next;
-        left.next.next = next;
-        left = left.next.next;
-    }
-}
-```
-
-<!-- tabs:end -->
-
 <!-- end -->