021_合并两个有序链表

简单

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        if(list1 == null){
            return list2;
        }
        if(list2 == null){
            return list1;
        }
        ListNode res = new ListNode();
        ListNode move = res;
        while(list1 != null || list2 != null){
            if(list1 == null){
                move.next = list2;
                break;
            }
            else if(list2 == null){
                move.next = list1;
                break;
            }
            else if(list1.val <= list2.val){
                move.next = list1;
                list1 = list1.next;
                move = move.next;
            }else{
                move.next = list2;
                list2 = list2.next;
                move = move.next;
            }
        }
        return res.next;
    }
}

Python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        res = ListNode()
        res_cur = res
        list1_cur = list1
        list2_cur = list2
        while(True):
            if(list1_cur == None):
                res_cur.next = list2_cur
                return res.next
            elif(list2_cur == None):
                res_cur.next = list1_cur
                return res.next
            elif(list1_cur.val <= list2_cur.val):
                res_cur.next = list1_cur
                res_cur = res_cur.next
                list1_cur = list1_cur.next
            else:
                res_cur.next = list2_cur
                res_cur = res_cur.next
                list2_cur = list2_cur.next
        return res.next

建立一个res链表用于返回，通过list1，list2两个节点遍历链表，将值赋给新的链表节点中