147 lines
2.9 KiB
Markdown
147 lines
2.9 KiB
Markdown
```c++
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void removeElements(LinkedNode* head, int val) {
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if (head == nullptr) {
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return;
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}
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LinkedNode* cur;
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LinkedNode* dummyHead = new LinkedNode(0); // 虚拟头节点
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dummyHead->next = head; // 将虚拟头节点连接到原链表
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cur = dummyHead; // 从虚拟头节点开始遍历
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while(cur->next != nullptr) {
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if(cur->next->val == val) {
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LinkedNode* tmp = cur->next;
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cur->next = cur->next->next;
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delete tmp;
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tmp = nullptr;
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}
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else {
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cur = cur->next;
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}
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}
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head = dummyHead->next;
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delete dummyHead;
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dummyHead = nullptr;
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}
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A. dummyHead->next = head; cur = dummyHead;
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B. dummyHead->next = head->next; cur = dummyHead;
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C. dummyHead->next = head; cur = dummyHead->next;
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D. dummyHead->next = head->next; cur = dummyHead->next;
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```
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这样可以正确处理所有情况，包括：
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- 删除普通节点
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- 删除头节点
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- 删除连续节点
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- 删除尾节点
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### 此题目的重占在于操作的是：cur->next
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我来用 ASCII 图解释这个过程，假设我们要删除值为 $3$ 的节点。
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原始链表：
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```
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1 -> 3 -> 3 -> 2 -> 3 -> 6
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```
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1 创建虚拟头节点并连接（选项 $A$）：
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```
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dummyHead(0) -> 1 -> 3 -> 3 -> 2 -> 3 -> 6
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^
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cur
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```
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2 遍历过程：
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第一步：$cur->next$ (值为1) ≠ 3，移动 $cur$
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```
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dummyHead(0) -> 1 -> 3 -> 3 -> 2 -> 3 -> 6
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^
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cur
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```
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第二步：$cur->next$ (值为3) = 3，删除节点
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```
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dummyHead(0) -> 1 -> 3 -> 3 -> 2 -> 3 -> 6
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^ |
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| ×
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cur |
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+----+
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```
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删除后：
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```
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dummyHead(0) -> 1 -> 3 -> 2 -> 3 -> 6
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^
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cur
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```
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继续类似操作，最终结果：
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```
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dummyHead(0) -> 1 -> 2 -> 6
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```
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如果是其他选项：
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选项 B (dummyHead->next = head->next; cur = dummyHead):
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```
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原始： 1 -> 3 -> 3 -> 2 -> 3 -> 6
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变成：
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dummyHead(0) -> 3 -> 3 -> 2 -> 3 -> 6 // 丢失了节点1
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^
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cur
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```
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选项 C (dummyHead->next = head; cur = dummyHead->next):
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```
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dummyHead(0) -> 1 -> 3 -> 3 -> 2 -> 3 -> 6
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^
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cur // cur 从1开始，无法处理第一个节点
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```
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选项 D (dummyHead->next = head->next; cur = dummyHead->next):
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```
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dummyHead(0) -> 3 -> 3 -> 2 -> 3 -> 6
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^
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cur // 既丢失了节点1，又无法处理第一个节点
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```
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所以选项 $A$ 是正确的，因为：
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保留了完整的原始链表
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cur 从虚拟头节点开始，可以处理所有节点
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通过 cur->next 来判断和删除节点，操作更安全 |