【LeetCode】Merge Two Sorted Lists

Merge Two Sorted Lists

Total Accepted: 18308 Total Submissions: 55982 My Submissions

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.

【解题思路】

归并排序，考察基本功。

赋值的时候可以new一个节点，也可以将一个链表赋值给另外一个，后者相对慢一些。

Java AC 440ms 新建节点

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode node = new ListNode(0);
        ListNode point = node;
        while(l1 != null && l2 != null){
            if(l1.val > l2.val){
                point.next = new ListNode(l2.val);
                point = point.next;
                l2 = l2.next;
            }else{
                point.next = new ListNode(l1.val);
                point = point.next;
                l1 = l1.next;
            }
        }
        while(l1 != null){
            point.next = new ListNode(l1.val);
            point = point.next;
            l1 = l1.next;
        }
        while(l2 != null){
            point.next = new ListNode(l2.val);
            point = point.next;
            l2 = l2.next;
        }
        return node.next;
    }
}

Python AC 224ms 新建节点

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param two ListNodes
    # @return a ListNode
    def mergeTwoLists(self, l1, l2):
        node = ListNode(0)
        point = node
        while l1 is not None and l2 is not None:
            if l1.val > l2.val:
                point.next = ListNode(l2.val)
                l2 = l2.next
            else:
                point.next = ListNode(l1.val)
                l1 = l1.next
            point = point.next
        while l1 is not None:
            point.next = ListNode(l1.val)
            point = point.next
            l1 = l1.next
        while l2 is not None:
            point.next = ListNode(l2.val)
            point = point.next
            l2 = l2.next
        return node.next

Java AC 576ms 节点赋值

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
	public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
		ListNode node = new ListNode(0);
		ListNode point = node;
		while (l1 != null && l2 != null) {
			if (l1.val > l2.val) {
				point.next = l2;
				point = point.next;
				l2 = l2.next;
			} else {
				point.next = l1;
				point = point.next;
				l1 = l1.next;
			}
		}
		if (l1 != null) {
			point.next = l1;
		}
		if (l2 != null) {
			point.next = l2;
		}
		return node.next;
	}
}

Python AC 240ms 节点赋值

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param two ListNodes
    # @return a ListNode
    def mergeTwoLists(self, l1, l2):
        node = ListNode(0)
        point = node
        while l1 is not None and l2 is not None:
            if l1.val > l2.val:
                point.next = l2
                l2 = l2.next
            else:
                point.next = l1
                l1 = l1.next
            point = point.next
        if l1 is not None:
            point.next = l1
        if l2 is not None:
            point.next = l2
        return node.next