问题描述
输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};
方法一
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
ListNode* newList = NULL; //新链表头
ListNode* newListRear = NULL; //新链表尾
// 先处理某个链表为空的情形
if (pHead1 == NULL){
return pHead2;
}
if (pHead2 == NULL){
return pHead1;
}
// 把数值小的结点放入新链表，生成头节点
if (pHead1->val <= pHead2->val){
newList = pHead1;
newListRear = pHead1;
pHead1 = pHead1->next;
}else{
newList = pHead2 ;
newListRear = pHead2;
pHead2 = pHead2->next;
}
// 两表均不空的情形下，遍历
while (pHead1 != NULL && pHead2 != NULL) {
if (pHead1->val <= pHead2->val) {
newListRear->next =pHead1;
newListRear = pHead1;
pHead1 = pHead1->next;
}else{
newListRear->next =pHead2;
newListRear = pHead2;
pHead2 = pHead2->next;
}
}
//某一表为空时，把另一表接入新表表尾
if (pHead1 == NULL) {
newListRear->next = pHead2;
}
if (pHead2 == NULL) {
newListRear->next = pHead1;
}
return newList;
}
};
方法二（递归思想）
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
if (pHead1 == NULL){
return pHead2;
}
if (pHead2 == NULL){
return pHead1;
}
if (pHead1->val <= pHead2->val){ // pHead1为合并后的头节点
pHead1->next = Merge(pHead1->next, pHead2);
return pHead1;
}else{ // pHead2 为合并后的头节点
pHead2->next = Merge(pHead1, pHead2->next);
return pHead2;
}
}
};
