剑指Offer面试题17 & Leetcode21

剑指Offer面试题17 & Leetcode21

解题思路

  考虑：链表1的头结点和链表2的头结点中的较小者为合并后链表的头结点，之后合并的是链表1第二个节点开始的链表和链表2，可以用递归的思想解决问题。注意处理空指针问题。当然也可以用循环，用一个ListNode保留头节点，然后用另一个ListNode遍历两个链表。

Solution1 递归

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if(l1 == null)
return l2;
if(l2 == null)
return l1;
ListNode head;
if(l1.val<l2.val){
head = l1;
head.next = mergeTwoLists(l1.next, l2);
}else{
head = l2;
head.next = mergeTwoLists(l1, l2.next);
}
return head;
}

Solution2 循环

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) {
return l2;
}
if (l2 == null) {
return l1;
}
ListNode node = null;
ListNode head = null;
while (l1 != null && l2 != null) {
if (l1.val <= l2.val) {
if (node == null) {
node = l1;
head = node;
} else {
node.next = l1;
node = node.next;
}
l1 = l1.next;
} else {
if (node == null) {
node = l2;
head = node;
} else {
node.next = l2;
node = node.next;
}
l2 = l2.next;
}
}
if (l1 != null) {
node.next = l1;
} else if (l2 != null) {
node.next = l2;
}
return head;
}