![[LeetCode] 21. Merge Two Sorted Lists 解题思路 [LeetCode] 21. Merge Two Sorted Lists 解题思路](https://image.shishitao.com:8440/aHR0cHM6Ly9ia3FzaW1nLmlrYWZhbi5jb20vdXBsb2FkL2NoYXRncHQtcy5wbmc%2FIQ%3D%3D.png?!?w=700&webp=1)
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.
问题:将两个已排序的列表,合并为一个有序列表。
令 head 为两个列表表头中较小的一个,令 p 为新的已排序的最后一个元素。令 l1, l2 分别为两个列表中未排序部分的首节点。依次将 l1, l2 中的较小值追加到 p 后面,并调整 p 和 l1、l2较小者指针即可。
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { if (l1 == NULL) {
return l2;
} if (l2 == NULL) {
return l1;
} ListNode* head = new ListNode();
if (l1->val <= l2->val) {
head = l1;
l1 = l1->next;
}else{
head = l2;
l2 = l2->next;
} ListNode* p = head; while(l1 != NULL && l2 != NULL){ if(l1->val <= l2->val){
p->next = l1;
p = p->next;
l1 = l1->next;
}else{
p->next = l2;
p = p->next;
l2 = l2->next;
}
} if (l1 == NULL) {
p->next = l2;
} if (l2 == NULL) {
p->next = l1;
} return head;
}