[LeetCode]80. Merge Two Sorted Lists合并两个排序链表

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.

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解法1：递归。首先比较头节点大小，若l2->val>l1->val，则返回mergeTwoLists(l2,l1)；否则（1）如果l1->next!=NULL，则比较l1->next->val与l2->val的大小，这样可以确定前两个元素的顺序，

/**
* 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;
if (l2 == NULL) return l1;
if (l1->val <= l2->val) {
if (l1->next != NULL && l1->next->val <= l2->val){
ListNode* node = mergeTwoLists(l1->next->next, l2);
l1->next->next = node;
}
else if (l1->next != NULL && l1->next->val > l2->val) {
ListNode* node = mergeTwoLists(l1->next, l2->next);
l2->next = node;
l1->next = l2;
}
else
l1->next = l2;
return l1;
}
else
return mergeTwoLists(l2, l1);
}
};

解法2：迭代。新建一个指针help，然后遍历l1和l2，将二者中较小的链接到help后面。最后返回help的下一个节点作为头节点。注意两个链表可能长度不一样，要将长链表最后剩下的元素链接到新链表中。

/**
* 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;
if (l2 == NULL) return l1;
ListNode* help = new ListNode(0);
ListNode* head = help;
while (l1 != NULL && l2 != NULL) {
if (l1->val <= l2->val) {
help->next = l1;
l1 = l1->next;
}
else {
help->next = l2;
l2 = l2->next;
}
help = help->next;
}
if (l1 != NULL) help->next = l1;
if (l2 != NULL) help->next = l2;
return head->next;
}
};

转载于:https://www.cnblogs.com/aprilcheny/p/4964996.html