leetcode 21 合并两个有序链表

简介

用了一个比较复杂的方法, 直接讲两个链表的数据取出来,使用vector进行排序, 然后再链接两个链表, 在把数据插回去. 不如递归简单.

code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode *p = l1;
        ListNode *q = l2;
        if(p == nullptr && q == nullptr){
            return p;
        }
        // 将p插入q中
        vector<int> v;
        while(p){
            v.push_back(p->val);
            p = p->next;
        }
        while(q){
            v.push_back(q->val);
            q = q->next;
        }
        sort(v.begin(), v.end());
        if(l1 != nullptr){
             p = l1;
             while(p->next){
                p = p->next;
            }
            p->next = l2;
            p = l1;
            int i=0;
            while(p){
                p->val = v[i];
                i++;
                p = p->next; 
            }
            return l1;
        }else{
            q = l2;
             while(q->next){
                q = q->next;
            }
            q->next = l1;
            q = l2;
            int i=0;
            while(q){
                q->val = v[i];
                i++;
                q = q->next; 
            }
            return l2;
        }
        return nullptr;

        
    }
};

class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1 == null){
            return l2;
        } else if(l2 == null){
            return l1;
        } else if(l1.val < l2.val){
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        } else{
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }
}

TIPS

递归算法很巧妙, 不用构造零时指针. 及其方便