题目描述
将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
示例：
输入：1->2->4, 1->3->4 输出：1->1->2->3->4->4
解题思路
稍简单，不影响打dota暴走，咱们就过了。
代码
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
struct ListNode *pstHead;
struct ListNode *pstTmp;
if (0 == l1)
{
return l2;
}
if (0 == l2)
{
return l1;
}
if (l1->val > l2->val)
{
pstHead = l2;
l2 = l2->next;
}
else
{
pstHead = l1;
l1 = l1->next;
}
pstTmp = pstHead;
while (0 != l1 && 0 != l2)
{
if (l1->val > l2->val)
{
pstTmp->next = l2;
l2 = l2->next;
}
else
{
pstTmp->next = l1;
l1 = l1->next;
}
pstTmp = pstTmp->next;
}
if (0 == l1)
{
pstTmp->next = l2;
}
if (0 == l2)
{
pstTmp->next = l1;
}
return pstHead;
}
看答案
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2){
if(l1 == NULL)
return l2;
if(l2 == NULL)
return l1;
struct ListNode *head,*cur1,*cur2,*pre,*next;
if(l1->val < l2->val) {
head = l1;
} else {
head = l2;
}
if(l1->val < l2->val) {
cur1 = l1;
cur2 = l2;
} else {
cur1 = l2;
cur2 = l1;
}
while(cur1 != NULL && cur2 != NULL){
if(cur1->val <= cur2->val){
pre = cur1;
cur1 = cur1->next;
}else{
next = cur2->next;
pre->next = cur2;
cur2->next = cur1;
pre = cur2;
cur2 = next;
}
}
if(cur1 == NULL) {
pre -> next = cur2;
} else {
pre -> next = cur1;
}
return head;
}
总结
没有总结。
