面试题25. 合并两个排序的链表
思路:最令我困惑的是,几乎所有人都采取了减而治之的策略,最优解也是如此,但这题却分类在分而治之…
本题采用减而治之的策略,逐步蚕食问题的规模,最终可以得到问题的答案。不想解释了,简单题(对现在的我来说)。
迭代:
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if(!l1 || !l2) return l1 == NULL ? l2 : l1;
ListNode* visual = new ListNode(1);
ListNode* pre = visual;
while(l1 && l2)
{
if(l1->val <= l2->val)
{
pre->next = l1;
l1 = l1->next;
}
else
{
pre->next = l2;
l2 = l2->next;
}
pre = pre->next;
}
l1 == NULL ? pre->next = l2 : pre->next = l1;
pre = visual->next;
delete visual;
return prew;
}
};
递归:
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if (!l1 || !l2) return l1 == NULL ? l2 : l1;
if (l1->val <= l2->val)
{
l1->next = mergeTwoLists(l1->next, l2);
return l1;
}
l2->next = mergeTwoLists(l1, l2->next);
return l2;
}
};