reverse linked list

206.reverse-linked-list.md

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+## 题目地址
+https://leetcode.com/problems/reverse-linked-list/description/
+
+## 题目描述
+Reverse a singly linked list.
+
+Example:
+
+Input: 1->2->3->4->5->NULL
+Output: 5->4->3->2->1->NULL
+Follow up:
+
+A linked list can be reversed either iteratively or recursively. Could you implement both?
+
+## 思路
+这个就是常规操作了，使用一个变量记录前驱pre，一个变量记录后继next.
+
+不断更新`current.next = pre` 就好了
+## 关键点解析
+
+- 链表的基本操作（交换）
+- 虚拟节点dummy 简化操作
+- 注意更新current和pre的位置， 否则有可能出现溢出
+
+
+## 代码
+
+```js
+/*
+ * @lc app=leetcode id=206 lang=javascript
+ *
+ * [206] Reverse Linked List
+ *
+ * https://leetcode.com/problems/reverse-linked-list/description/
+ *
+ * algorithms
+ * Easy (52.95%)
+ * Total Accepted:    532.6K
+ * Total Submissions: 1M
+ * Testcase Example:  '[1,2,3,4,5]'
+ *
+ * Reverse a singly linked list.
+ * 
+ * Example:
+ * 
+ * 
+ * Input: 1->2->3->4->5->NULL
+ * Output: 5->4->3->2->1->NULL
+ * 
+ * 
+ * Follow up:
+ * 
+ * A linked list can be reversed either iteratively or recursively. Could you
+ * implement both?
+ * 
+ */
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {ListNode}
+ */
+var reverseList = function(head) {
+    const dummyHead = {
+        next: head
+    }
+    let current = dummyHead.next;
+    let pre = null;
+
+    while(current) {
+        const next = current.next;
+        current.next = pre;
+        pre = current;
+        current = next;
+    }
+    return pre;
+};
+
+```
+

92.reverse-linked-list-ii.md

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+## 题目地址
+https://leetcode.com/problems/reverse-linked-list-ii/description/
+
+## 题目描述
+Reverse a linked list from position m to n. Do it in one-pass.
+
+Note: 1 ≤ m ≤ n ≤ length of list.
+
+Example:
+
+Input: 1->2->3->4->5->NULL, m = 2, n = 4
+Output: 1->4->3->2->5->NULL
+
+## 思路
+
+考虑取出需要反转的这一小段链表，反转完后再插入到原先的链表中。
+
+以本题为例：
+
+变换的是2,3,4这三个点，那么我们可以先取出2，用front指针指向2，然后当取出3的时候，我们把3加到2的前面，把front指针前移到3，依次类推，到4后停止，这样我们得到一个新链表4->3->2, front指针指向4。
+
+对于原链表来说，有两个点的位置很重要，需要用指针记录下来，分别是1和5，把新链表插入的时候需要这两个点的位置。
+
+用pre指针记录1的位置
+
+当4结点被取走后，5的位置需要记下来
+
+这样我们就可以把倒置后的那一小段链表加入到原链表中
+
+![92.reverse-linked-list-ii](./assets/92.reverse-linked-list-ii.gif)
+
+(图片来自： https://github.com/MisterBooo/LeetCodeAnimation)
+## 关键点解析
+
+- 链表的基本操作（交换）
+- 虚拟节点dummy 简化操作
+- 考虑特殊情况 m 是 1 或者 n是链表长度的情况
+- 用四个变量记录特殊节点， 然后操作这四个节点使之按照一定方式连接即可。
+
+```js
+    let midStartNode = null;
+    let preMidStartNode = null;
+    let midEndNode = null;
+    let postMidEndNode = null;
+```
+
+- 注意更新current和pre的位置， 否则有可能出现溢出
+
+
+## 代码
+
+```js
+/*
+ * @lc app=leetcode id=92 lang=javascript
+ *
+ * [92] Reverse Linked List II
+ *
+ * https://leetcode.com/problems/reverse-linked-list-ii/description/
+ *
+ * algorithms
+ * Medium (34.13%)
+ * Total Accepted:    182.3K
+ * Total Submissions: 532.8K
+ * Testcase Example:  '[1,2,3,4,5]\n2\n4'
+ *
+ * Reverse a linked list from position m to n. Do it in one-pass.
+ * 
+ * Note: 1 ≤ m ≤ n ≤ length of list.
+ * 
+ * Example:
+ * 
+ * 
+ * Input: 1->2->3->4->5->NULL, m = 2, n = 4
+ * Output: 1->4->3->2->5->NULL
+ * 
+ * 
+ */
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @param {number} m
+ * @param {number} n
+ * @return {ListNode}
+ */
+var reverseBetween = function(head, m, n) {
+    const dummyHead = {
+        next: head
+    }
+
+    let current = dummyHead.next;
+    let pre = current;
+    let index = 0;
+
+    let midStartNode = null;
+    let preMidStartNode = null;
+    let midEndNode = null;
+    let postMidEndNode = null;
+
+    while(current) {
+        const next = current.next;
+        index++;
+
+        // reverse linked list
+        if (index > m && index <= n) {           
+            current.next = pre;
+        }
+
+        if (index === m - 1) {
+            preMidStartNode = current;
+        }
+        if (index === m) {
+            midStartNode = current;
+        }
+
+        if (index === n + 1) {
+            postMidEndNode = current;
+            postMidEndNode.next
+        }
+
+        if (index === n) {
+            midEndNode = current;;
+        }
+
+        pre = current;
+
+        current = next;
+    }
+
+    // 两个链表合并起来
+    (preMidStartNode || dummyHead).next = midEndNode; // 特殊情况需要考虑
+    midStartNode.next = postMidEndNode;
+
+    return dummyHead.next;
+};
+
+```