[LeetCode] Merge Two Sorted Lists

最新推荐文章于 2025-04-27 14:03:08 发布

最新推荐文章于 2025-04-27 14:03:08 发布 · 73 阅读

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本文提供了一种将两个已排序的单链表合并为一个新的排序链表的方法。通过比较两个链表当前节点的值，选择较小值的节点作为新链表的一部分，并递进到下一个节点，直至所有节点被处理。

刷题，练习手感！

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        if (l1 == NULL) return l2;
        if (l2 == NULL) return l1;
        ListNode* head, *cur, *tmp;
        head = getNext(l1, l2);
        cur = head;
        while (true) {
            if (l1 == NULL || l2 == NULL) break;
            cur->next = getNext(l1, l2);
            cur = cur->next;
        }
        if (l1 != NULL) cur->next = l1;
        if (l2 != NULL) cur->next = l2;
        return head;
    }
    
    ListNode* getNext(ListNode*& l1, ListNode*&l2) {
        if (l1->val > l2->val) {
            ListNode* t = l2;
            l2 = l2->next;
            return t;
        } else {
            ListNode* t = l1;
            l1 = l1->next; 
            return t;
        }
    }
};

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        if (l1 == NULL || l2 == NULL) return l1 == NULL ? l2 : l1;
        ListNode head(0);
        ListNode *cur = &head;
        while (l1 != NULL && l2 != NULL) {
            if (l1->val < l2->val) {
                cur->next = l1;
                l1 = l1->next;
                cur = cur->next;
            }
            else {
                cur->next= l2;
                l2 = l2->next;
                cur = cur->next;
            }
        }
        if (l1 != NULL) cur->next= l1;
        else if (l2 != NULL) cur->next = l2;
        return head.next;
    }
};