17 - Merge Two Sorted Lists

最新推荐文章于 2023-02-25 02:16:34 发布

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本文提供了一种将两个已排序的链表合并为一个新排序链表的方法。通过迭代或递归的方式，有效地实现了节点的拼接，确保新链表仍保持有序状态。

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

solution ：非常简单，小心遇到NULL即可。

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(l1 == NULL && l2 == NULL)
            return NULL;
        else if(l1 != NULL && l2 == NULL)
            return l1;
        else if(l1 == NULL && l2 != NULL)
            return l2;
        
        ListNode *newHead = new ListNode(0);
        ListNode *newNode = newHead;
        
        ListNode *node1 = l1;
        ListNode *node2 = l2;
        
        while(node1 != NULL && node2 != NULL)
        {
            int val1 = node1 -> val;    
            int val2 = node2 -> val;
            
            if(val1 < val2) //node1 be the target node
            {
                newNode -> next = node1;
                newNode = node1;
                node1 = node1 -> next;
                newNode -> next = NULL;
            }
            else //node2 be the target node
            {
                newNode -> next = node2;
                newNode = node2;
                node2 = node2 -> next;
                newNode -> next = NULL;
            }
            
        }
        
        while(node1 != NULL)
        {
            newNode -> next = node1;
            newNode = node1;
            node1 = node1 -> next;
            newNode -> next = NULL;
        }
        
        while(node2 != NULL)
        {
            newNode -> next = node2;
            newNode = node2;
            node2 = node2 -> next;
            newNode -> next = NULL;
        }
        
        newHead = newHead -> next;
        
        return newHead;
    }
};

上面的代码还是太长了，面试的时候还是简洁的代码为主，也就是说能递归的尽量递归：

class Solution {
public:
ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
    // Start typing your C/C++ solution below
    // DO NOT write int main() function
    if (l1==NULL) return l2;
    if (l2==NULL) return l1;
    ListNode *ret;
    if (l1->val < l2->val) {
        ret = l1;
        ret->next = mergeTwoLists(l1->next,l2);
    }  else {
        ret = l2;
        ret->next = mergeTwoLists(l1,l2->next);
    }
    return ret;
}
};