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21. Merge Two Sorted Lists
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.
Example:
Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4
思路分析
最先想到的是迭代。
但是，也可以用递归解决！
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一刷
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三刷
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public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (Objects.isNull(l1) && Objects.isNull(l2)) {
return null;
}
ListNode dummy = new ListNode();
ListNode tail = dummy;
while (Objects.nonNull(l1) && Objects.nonNull(l2)) {
if (l1.val < l2.val) {
tail.next = l1;
l1 = l1.next;
} else {
tail.next = l2;
l2 = l2.next;
}
tail = tail.next;
}
if (Objects.isNull(l1)) {
tail.next = l2;
}
if (Objects.isNull(l2)) {
tail.next = l1;
}
return dummy.next;
}
/**
* Runtime: 1 ms, faster than 28.15% of Java online submissions for Merge Two Sorted Lists.
*
* Memory Usage: 39.8 MB, less than 14.45% of Java online submissions for Merge Two Sorted Lists.
*/
public ListNode mergeTwoLists1(ListNode l1, ListNode l2) {
if (Objects.isNull(l1)) {
return l2;
}
if (Objects.isNull(l2)) {
return l1;
}
ListNode result = null;
ListNode p1 = l1;
ListNode p2 = l2;
ListNode tail = null;
while (Objects.nonNull(p1) && Objects.nonNull(p2)) {
int v1 = p1.val;
int v2 = p2.val;
ListNode temp = null;
if (v1 < v2) {
temp = p1;
p1 = p1.next;
} else {
temp = p2;
p2 = p2.next;
}
if (Objects.isNull(tail)) {
result = temp;
tail = temp;
} else {
tail.next = temp;
tail = temp;
}
}
if (Objects.isNull(p1)) {
tail.next = p2;
}
if (Objects.isNull(p2)) {
tail.next = p1;
}
return result;
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-07-02 19:43:59
*/
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (Objects.isNull(l1)) {
return l2;
}
if (Objects.isNull(l2)) {
return l1;
}
ListNode dummy = new ListNode();
ListNode curr = dummy;
while (l1 != null && l2 != null) {
if (l1.val <= l2.val) {
curr.next = l1;
l1 = l1.next;
} else if (l2.val < l1.val) {
curr.next = l2;
l2 = l2.next;
}
curr = curr.next;
if (l1 == null) {
curr.next = l2;
break;
}
if (l2 == null) {
curr.next = l1;
break;
}
}
return dummy.next;
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-07-02 19:43:59
*/
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) {
return l2;
} else if (l2 == null) {
return l1;
} else if (l1.val < l2.val) {
// 升序链表，则把最小的留下，其余的进行递归。
l1.next = mergeTwoLists(l1.next, l2);
return l1;
} else {
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
}