Merge pull request #188 from LetMeFly666/82

添加了问题“82.删除排序链表中的重复元素II”的代码和题解

Author: LetMeFly666

Codes/0082-remove-duplicates-from-sorted-list-ii.cpp

@@ -0,0 +1,31 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2024-01-15 21:33:23
+ * @LastEditors: LetMeFly
+ * @LastEditTime: 2024-01-15 21:43:18
+ */
+#ifdef _WIN32
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+public:
+    ListNode* deleteDuplicates(ListNode* head) {
+        ListNode* ans = new ListNode(1000, head);
+        ListNode* lastNode = ans, *thisNode = head;
+        while (thisNode && thisNode->next) {
+            if (thisNode->val == thisNode->next->val) {
+                ListNode* nextNode = thisNode->next->next;
+                while (nextNode && thisNode->val == nextNode->val) {
+                    nextNode = nextNode->next;
+                }
+                lastNode->next = nextNode;
+                thisNode = nextNode;
+            }
+            else {
+                lastNode = thisNode, thisNode = thisNode->next;
+            }
+        }
+        return ans->next;
+    }
+};

Codes/0082-remove-duplicates-from-sorted-list-ii.py

@@ -0,0 +1,28 @@
+'''
+Author: LetMeFly
+Date: 2024-01-15 21:45:39
+LastEditors: LetMeFly
+LastEditTime: 2024-01-15 21:53:04
+'''
+from typing import Optional
+
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+class Solution:
+    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        ans = ListNode(1000, head)
+        lastNode, thisNode = ans, head
+        while thisNode and thisNode.next:
+            if thisNode.val == thisNode.next.val:
+                nextNode = thisNode.next.next
+                while nextNode and thisNode.val == nextNode.val:
+                    nextNode = nextNode.next
+                lastNode.next = nextNode
+                thisNode = nextNode
+            else:
+                lastNode, thisNode = thisNode, thisNode.next
+        return ans.next

README.md

@@ -54,6 +54,7 @@
 |0062.不同路径|中等|<a href="https://leetcode.cn/problems/unique-paths/solutions/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2022/11/26/LeetCode%200062.%E4%B8%8D%E5%90%8C%E8%B7%AF%E5%BE%84/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/128059055" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/unique-paths/solutions/1992721/letmefly-62bu-tong-lu-jing-liang-chong-f-e8tl/" target="_blank">LeetCode题解</a>|
 |0067.二进制求和|简单|<a href="https://leetcode.cn/problems/add-binary/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2022/07/14/LeetCode%200067.%E4%BA%8C%E8%BF%9B%E5%88%B6%E6%B1%82%E5%92%8C/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/125793685" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/add-binary/solution/by-tisfy-o5z5/" target="_blank">LeetCode题解</a>|
 |0070.爬楼梯|简单|<a href="https://leetcode.cn/problems/climbing-stairs/solutions/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2023/12/10/LeetCode%200070.%E7%88%AC%E6%A5%BC%E6%A2%AF/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/134913892" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/climbing-stairs/solutions/2561558/letmefly-70pa-lou-ti-dong-tai-gui-hua-di-7d8m/" target="_blank">LeetCode题解</a>|
+|0082.删除排序链表中的重复元素II|中等|<a href="https://leetcode.cn/problems/remove-duplicates-from-sorted-list-ii/solutions/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2024/01/15/LeetCode%200082.%E5%88%A0%E9%99%A4%E6%8E%92%E5%BA%8F%E9%93%BE%E8%A1%A8%E4%B8%AD%E7%9A%84%E9%87%8D%E5%A4%8D%E5%85%83%E7%B4%A0II/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/135612345" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/remove-duplicates-from-sorted-list-ii/solutions/2605736/letmefly-82shan-chu-pai-xu-lian-biao-zho-92xm/" target="_blank">LeetCode题解</a>|
 |0083.删除排序链表中的重复元素|简单|<a href="https://leetcode.cn/problems/remove-duplicates-from-sorted-list/solutions/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2024/01/14/LeetCode%200083.%E5%88%A0%E9%99%A4%E6%8E%92%E5%BA%8F%E9%93%BE%E8%A1%A8%E4%B8%AD%E7%9A%84%E9%87%8D%E5%A4%8D%E5%85%83%E7%B4%A0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/135581000" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/remove-duplicates-from-sorted-list/solutions/2603584/letmefly-83shan-chu-pai-xu-lian-biao-zho-cey8/" target="_blank">LeetCode题解</a>|
 |0086.分隔链表|中等|<a href="https://leetcode.cn/problems/partition-list/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2022/06/26/LeetCode%200086.%E5%88%86%E9%9A%94%E9%93%BE%E8%A1%A8/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/125467594" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/partition-list/solution/letmefly-86fen-ge-lian-biao-by-tisfy-64z3/" target="_blank">LeetCode题解</a>|
 |0088.合并两个有序数组|简单|<a href="https://leetcode.cn/problems/merge-sorted-array/solutions/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2023/08/13/LeetCode%200088.%E5%90%88%E5%B9%B6%E4%B8%A4%E4%B8%AA%E6%9C%89%E5%BA%8F%E6%95%B0%E7%BB%84/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/132256535" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/merge-sorted-array/solutions/2385633/letmefly-88he-bing-liang-ge-you-xu-shu-z-u9cc/" target="_blank">LeetCode题解</a>|

Solutions/LeetCode 0082.删除排序链表中的重复元素II.md

@@ -0,0 +1,111 @@
+---
+title: 82.删除排序链表中的重复元素 II
+date: 2024-01-15 21:42:35
+tags: [题解, LeetCode, 中等, 链表, 双指针]
+---
+
+# 【LetMeFly】82.删除排序链表中的重复元素 II：模拟
+
+力扣题目链接：[https://leetcode.cn/problems/remove-duplicates-from-sorted-list-ii/](https://leetcode.cn/problems/remove-duplicates-from-sorted-list-ii/)
+
+<p>给定一个已排序的链表的头&nbsp;<code>head</code> ，&nbsp;<em>删除原始链表中所有重复数字的节点，只留下不同的数字</em>&nbsp;。返回 <em>已排序的链表</em>&nbsp;。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/01/04/linkedlist1.jpg" style="height: 142px; width: 500px;" />
+<pre>
+<strong>输入：</strong>head = [1,2,3,3,4,4,5]
+<strong>输出：</strong>[1,2,5]
+</pre>
+
+<p><strong>示例 2：</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/01/04/linkedlist2.jpg" style="height: 164px; width: 400px;" />
+<pre>
+<strong>输入：</strong>head = [1,1,1,2,3]
+<strong>输出：</strong>[2,3]
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li>链表中节点数目在范围 <code>[0, 300]</code> 内</li>
+	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
+	<li>题目数据保证链表已经按升序 <strong>排列</strong></li>
+</ul>
+
+
+    
+## 方法一：模拟
+
+相同的节点可能被全部删除（头节点可能也会被删），因此我们可以新建一个“空的头节点ans”，ans的next指向head。
+
+使用两个节点lastNode和thisNode，lastNode指向上一个节点（防止当前遍历到的节点被删除），thisNode指向当前处理到的节点。当thisNode和thisNode.next都非空时：
+
++ 如果```thisNode.val == thisNode.next.val```，新建一个nextNode节点指向thisNode.next.next（最终指向第一个和thisNode的值不同的节点）。当nextNode非空且```nextNode.val == thisNode.val```时，nextNode不断后移。最后将lastNode.next赋值为nextNode，并将thisNode赋值为nextNode（删掉了中间具有相同元素的节点）。
++ 否则，将lastNode和thisNode分别赋值为thisNode和thisNode.next（相当于指针后移）
+
+最终返回“假头节点”ans的next即可。
+
++ 时间复杂度$O(len(listnode))$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    ListNode* deleteDuplicates(ListNode* head) {
+        ListNode* ans = new ListNode(1000, head);
+        ListNode* lastNode = ans, *thisNode = head;
+        while (thisNode && thisNode->next) {
+            if (thisNode->val == thisNode->next->val) {
+                ListNode* nextNode = thisNode->next->next;
+                while (nextNode && thisNode->val == nextNode->val) {
+                    nextNode = nextNode->next;
+                }
+                lastNode->next = nextNode;
+                thisNode = nextNode;
+            }
+            else {
+                lastNode = thisNode, thisNode = thisNode->next;
+            }
+        }
+        return ans->next;
+    }
+};
+```
+
+#### Python
+
+```python
+# from typing import Optional
+
+# # Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+
+class Solution:
+    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        ans = ListNode(1000, head)
+        lastNode, thisNode = ans, head
+        while thisNode and thisNode.next:
+            if thisNode.val == thisNode.next.val:
+                nextNode = thisNode.next.next
+                while nextNode and thisNode.val == nextNode.val:
+                    nextNode = nextNode.next
+                lastNode.next = nextNode
+                thisNode = nextNode
+            else:
+                lastNode, thisNode = thisNode, thisNode.next
+        return ans.next
+```
+
+> 同步发文于CSDN，原创不易，转载经作者同意后请附上[原文链接](https://blog.tisfy.eu.org/2024/01/15/LeetCode%200082.%E5%88%A0%E9%99%A4%E6%8E%92%E5%BA%8F%E9%93%BE%E8%A1%A8%E4%B8%AD%E7%9A%84%E9%87%8D%E5%A4%8D%E5%85%83%E7%B4%A0II/)哦~
+> Tisfy：[https://letmefly.blog.csdn.net/article/details/135612345](https://letmefly.blog.csdn.net/article/details/135612345)