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递归算法：合并与反转链表的艺术

news 2026/4/20 2:35:24

合并两个有序链表

合并两个有序链表是将两个升序排列的链表合并成一个新的升序链表。使用递归方法时，核心思路是：比较两个链表的头节点值，选择较小的节点作为新链表的头，然后递归地合并剩余部分。如果其中一个链表为空，直接返回另一个链表作为结果。

递归过程：

如果链表1为空，返回链表2。
如果链表2为空，返回链表1。
否则，比较链表1和链表2的头节点值：
- 如果链表1的头节点值较小，则将链表1的头节点作为新头，递归合并链表1的剩余部分和链表2。
- 否则，将链表2的头节点作为新头，递归合并链表1和链表2的剩余部分。

C++代码实现：

struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(nullptr) {} }; ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { if (l1 == nullptr) return l2; // 链表1为空，返回链表2 if (l2 == nullptr) return l1; // 链表2为空，返回链表1 if (l1->val < l2->val) { l1->next = mergeTwoLists(l1->next, l2); // 递归合并剩余部分 return l1; } else { l2->next = mergeTwoLists(l1, l2->next); // 递归合并剩余部分 return l2; } }

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反转链表

反转链表是将链表节点的顺序完全倒置。使用递归方法时，核心思路是：递归到链表的末尾（最后一个节点），然后从后向前调整指针方向。具体步骤是：递归调用直到当前节点的下一个节点为空（即尾节点），此时该节点成为新链表的头节点；然后逐层返回，修改指针指向。

递归过程：

递归终止条件：如果当前节点为空或下一个节点为空（即尾节点），返回当前节点作为新头。
递归调用：对当前节点的下一个节点进行递归，得到反转后的新链表。
调整指针：将当前节点的下一个节点的指针指向当前节点，实现反转。
清除当前节点的原始指针，避免循环引用。

C++代码实现：

ListNode* reverseList(ListNode* head) { if (head == nullptr || head->next == nullptr) { return head; // 终止条件：空节点或尾节点，返回头节点 } ListNode* newHead = reverseList(head->next); // 递归到末尾 head->next->next = head; // 调整指针，反转方向 head->next = nullptr; // 清除原始指针 return newHead; // 返回新头节点 }