LeetCode链表：多指针和虚拟头结点

链表操作的时间复杂度

操作	时间复杂度
查找元素	O(n)
插入元素	O(1)
删除元素	O(1)

链表的解题技巧

• 多指针
  • 2个，快慢指针
  • 3个，链表反转
  • 4个，结点交换
• 虚拟头结点
  • 常用于删除结点的情况
  • 其他情况也可具体分析

LeetCode 206

 给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。

 示例 1：
 输入：head = [1,2,3,4,5]
 输出：[5,4,3,2,1]

C++

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode *pre = nullptr;
        ListNode *current = head;
        while (current != nullptr) {
            ListNode *next = current->next;
            
            // 反转
            current->next = pre;
            
            // 调整指针，准备进行下一次反转
            pre = current;
            current = next;
        }
        
        return pre;
    }
};

Swift

class Solution {
    func removeElements(_ head: ListNode?, _ val: Int) -> ListNode? {
        let dummyHead = ListNode()
        dummyHead.next = head
        
        var pre: ListNode? = dummyHead
        var current: ListNode? = head
        
        while current != nil {
            let next = current?.next
            
            if current?.val == val {
                pre?.next = next    // 移除current
            } else {
                pre = current
            }
            
            current = next
        }
        
        return dummyHead.next
    }
}

LeetCode 203

 给你一个链表的头节点 head 和一个整数 val ，请你删除链表中所有满足 Node.val == val 的节点，并返回 新的头节点 。

 示例 1：
 输入：head = [1,2,6,3,4,5,6], val = 6
 输出：[1,2,3,4,5]

C++

class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode *dummyHead = new ListNode(0);
        dummyHead->next = head;
        
        ListNode *pre = dummyHead;
        ListNode *current = head;
        
        while (current != nullptr) {
            ListNode *next = current->next;
            
            if (current->val == val) {
                current->next = nullptr;
                pre->next = next;
                
                delete current;
            } else {
                pre = current;
            }
            
            current = next;
        }
        
        head = dummyHead->next;
        delete dummyHead;
        
        return head;
    }
};

Swift

class Solution {
    func removeElements(_ head: ListNode?, _ val: Int) -> ListNode? {
        let dummyHead = ListNode()
        dummyHead.next = head
        
        var pre: ListNode? = dummyHead
        var current: ListNode? = head
        
        while current != nil {
            let next = current?.next
            
            if current?.val == val {
                pre?.next = next    // 移除current
            } else {
                pre = current
            }
            
            current = next
        }
        
        return dummyHead.next
    }
}

更多解题代码查看GitHub

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