合并两个有序链表
LeetCode 21递归(要开栈,占空间)
/**
* 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 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;
}
}
}
迭代
设置一个dummy做头,接下来操作链表
/**
* 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 l1, ListNode l2) {
ListNode dummy = new ListNode(-1);
ListNode curNode = dummy;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
curNode.next = l1;
l1 = l1.next;
} else {
curNode.next = l2;
l2 = l2.next;
}
curNode = curNode.next;
}
curNode.next = l1 == null ? l2 : l1;
return dummy.next;
}
}
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