21. 合并两个有序链表

21. 合并两个有序链表

做法1：

构建虚拟头节点，而后双指针做法。

/**
 * 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* list1, ListNode* list2) {
        ListNode* head = new ListNode(-1);
        ListNode* cur = head;

        while(list1 != nullptr && list2 != nullptr){
            if(list1->val <= list2->val){
                cur->next = list1;
                cur = cur->next;
                list1 = list1->next;
            }
            else{
                cur->next = list2;
                cur = cur->next;
                list2 = list2->next;
            }
        }
        while(list1 != nullptr){
            cur->next = list1;
            cur = cur->next;
            list1 = list1->next;
        }
        while(list2 != nullptr){
            cur->next = list2;
            cur = cur->next;
            list2 = list2->next;
        }
        return head->next;
    }
};

做法2：

迭代。那么最终返回的是哪一个节点呢？是两个链表最开始节点中小的那个节点。

/**
 * 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* list1, ListNode* list2) {
        if(list1 == nullptr) return list2;
        if(list2 == nullptr) return list1;

        if(list1->val <= list2->val){
           list1->next = mergeTwoLists(list1->next, list2);
           return list1;
        }
        list2->next = mergeTwoLists(list1, list2->next);
        return list2;

    }
};