C語言合并兩個有序鏈表的方法：拼接指定的兩個有序鏈表的所有節(jié)點即可。例如兩個有序鏈表分別為【1->2->4】和【1->3->4】，合并后的有序鏈表為【1->1->2->3->4->4】。

具體方法：

將兩個有序鏈表合并為一個新的有序鏈表并返回。新鏈表是通過拼接給定的兩個鏈表的所有節(jié)點組成的。

輸入：

1->2->4, 1->3->4

輸出：

1->1->2->3->4->4

分析：兩個鏈表為有序鏈表，所以依次遍歷兩個鏈表比較大小即可。

代碼實現(xiàn)：

/**

?* Definition for singly-linked list.

?* struct ListNode {

?*???? int val;

?*???? struct ListNode *next;

?* };

?*/

struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2){

????if(l1==NULL){

????????return l2;

????}

????if(l2==NULL){

????????return l1;

????}

????struct ListNode *l = (struct ListNode*)malloc(sizeof(struct ListNode));

????l->next = NULL;

????struct ListNode *list1 = l1;

????struct ListNode *list2 = l2;

????if(l1->valval){

????????l->val=l1->val;

????????if(list1->next==NULL){

????????????l->next=list2;

????????????return l;

????????}

????????list1=list1->next;

????}else{

????????l->val=l2->val;

????????if(list2->next==NULL){

????????????l->next=list1;

????????????return l;

????????}

????????list2=list2->next;

????}

????struct ListNode *list = l;

????while(list1->next!=NULL&&list2->next!=NULL){

????????if(list1->val<=list2->val){

????????????struct ListNode *body = (struct ListNode *)malloc(sizeof(struct ListNode));

????????????body->val = list1->val;

????????????body->next = NULL;

????????????list->next = body;

????????????list = list->next;

????????????list1 = list1->next;

????????}else{

????????????struct ListNode *body = (struct ListNode*)malloc(sizeof(struct ListNode));

????????????body->val=list2->val;

????????????body->next=NULL;

????????????list->next=body;

????????????list=list->next;

????????????list2=list2->next;

????????}

????}

????if(list1->next==NULL){

????????while(list2->next!=NULL){

????????????if(list1->val<=list2->val){

????????????????list->next = list1;

????????????????list = list->next;

????????????????list->next=list2;

????????????????return l;

????????????}else{

????????????????struct ListNode *body = (struct ListNode*)malloc(sizeof(struct ListNode));

????????????????body->val=list2->val;

????????????????body->next=NULL;

????????????????list->next=body;

????????????????list=list->next;

????????????????list2=list2->next;

????????????}

????????}

????}else{

????????while(list1->next!=NULL){

????????????if(list2->val<=list1->val){

????????????????list->next=list2;

????????????????list=list->next;

????????????????list->next=list1;

????????????????return l;

????????????}else{

????????????????struct ListNode *body = (struct ListNode*)malloc(sizeof(struct ListNode));

????????????????body->val=list1->val;

????????????????body->next=NULL;

????????????????list->next=body;

????????????????list=list->next;

????????????????list1=list1->next;

????????????}

????????}

????}

????if(list1->next==NULL&&list2->next==NULL){

????????if(list1->val<=list2->val){

????????????list->next = list1;

????????????list=list->next;

????????????list->next=list2;

????????}else{

????????????list->next=list2;

????????????list=list->next;

????????????list->next=list1;

????????}

????}

????return l;

}

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