本文实例为大家分享了C++合并两个排序的链表,供大家参考,具体内容如下
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。
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struct ListNode {
int val;
struct ListNode *next;
ListNode( int x) :
val(x), next(NULL) {
}
};
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方法一
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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;
}
};
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方法二(递归思想)
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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;
}
}
};
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以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持服务器之家。
原文链接:https://blog.csdn.net/francis_xd/article/details/82803905