传送门
nowcoder
leetcode
题目描述
输入两个递增的链表，单个链表的长度为n，
合并这两个链表并使新链表中的节点仍然是递增排序的。
数据范围：0 <= n <= 1000，-1000 <= 节点数 <= 1000
要求：空间复杂度O(1)，时间复杂度O(n)
C++ 代码 - nowcoder
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class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
if (pHead1 == nullptr) return pHead2;
if (pHead2 == nullptr) return pHead1;
if (pHead1->val > pHead2->val) swap(pHead1, pHead2);
pHead1->next = Merge(pHead1->next, pHead2);
return pHead1;
}
};
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
if (pHead1 == nullptr) return pHead2;
if (pHead2 == nullptr) return pHead1;
ListNode *dummy = new ListNode(-1), *p = dummy;
while (pHead1 != nullptr && pHead2 != nullptr) {
if (pHead1->val > pHead2->val) swap(pHead1, pHead2);
p->next = pHead1;
pHead1 = pHead1->next;
p = p->next;
}
p->next = (pHead1 ? pHead1 : pHead2);
return dummy->next;
}
};
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C++ 代码 - leetcode
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class Solution {
public:
ListNode* trainningPlan(ListNode* l1, ListNode* l2) {
ListNode* dummy = new ListNode(-1);
ListNode *p = dummy, *p1 = l1, *p2 = l2;
while (p1 != nullptr && p2 != nullptr) {
if (p1->val <= p2->val) {
p->next = p1;
p1 = p1->next;
} else {
p->next = p2;
p2 = p2->next;
}
p = p->next;
}
p->next = (p1 != nullptr) ? p1 : p2;
return dummy->next;
}
};
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