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How to merge two ordered linked lists in C language

王林
Release: 2020-11-03 09:57:56
Original
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How to merge two ordered linked lists in C language: just splice all the nodes of the two specified ordered linked lists. For example, the two ordered linked lists are [1->2->4] and [1->3->4], and the merged ordered linked list is [1->1->2- >3->4->4].

How to merge two ordered linked lists in C language

Specific method:

Merge two ordered linked lists into a new ordered linked list and return. The new linked list is formed by concatenating all the nodes of the two given linked lists.

(Video tutorial recommendation: java course)

Input:

1->2->4, 1->3->4
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Output:

1->1->2->3->4->4
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Analysis: The two linked lists are Ordered linked lists, so just traverse the two linked lists in sequence to compare their sizes.

Code implementation:

/**
 * 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->val<l2->val){
        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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