题目
C++ solution
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if (l1 == NULL) {
return l2;
} else if (l2 == NULL) {
return l1;
}
int min = 0;
if (l1->val < l2->val) {
min = l1->val;
l1 = l1->next;
} else {
min = l2->val;
l2 = l2->next;
}
ListNode* result = new ListNode(min);
ListNode* temp = result;
while (l1 != NULL) {
if (l2 == NULL) {
temp->next = new ListNode(l1->val);
temp = temp->next;
l1 = l1->next;
continue;
}
if (l1->val < l2->val) {
min = l1->val;
l1 = l1->next;
} else {
min = l2->val;
l2 = l2->next;
}
temp->next = new ListNode(min);
temp = temp->next;
}
while (l2 != NULL) {
temp->next = new ListNode(l2->val);
temp = temp->next;
l2 = l2->next;
}
return result;
}
};
简要题解
合并两个有序的链表:
- 若其中一个链表为空,则返回另一个链表。
- 由于两个链表都是有序的,只要依次比较两个链表结点的值,将较小者加入结果链表中,直到遍历完两个链表的所有结点。