Leetcode—445.两数相加II【中等】

2023每日刷题（六十七）

Leetcode—445.两数相加II

实现代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */


struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
    struct ListNode* head1 = (struct ListNode*)malloc(sizeof(struct ListNode));
    head1->next = NULL;
    struct ListNode* head2 = (struct ListNode*)malloc(sizeof(struct ListNode));
    head2->next = NULL;
    struct ListNode* p, *q, *p2, *q2;
    p = l1;
    p2 = l1->next;
    q = l2;
    q2 = l2->next;
    // 链表翻转
    while(p != NULL) {
        p->next = head1->next;
        head1->next = p;
        p = p2;
        if(p2 == NULL) {
            break;
        }
        p2 = p2->next;
        
    }
    while(q != NULL) {
        q->next = head2->next;
        head2->next = q;
        q = q2;
        if(q2 == NULL) {
            break;
        }
        q2 = q2->next;
        
    }
    p = head1->next;
    q = head2->next;
    head1->next = NULL;
    int c = 0;
    while(p && q) {
        struct ListNode* s = (struct ListNode*)malloc(sizeof(struct ListNode));
        s->next = NULL;
        int sum = p->val + q->val + c;
        c = sum / 10;
        s->val = sum % 10;
        s->next = head1->next;
        head1->next = s;
        p = p->next;
        q = q->next;
    }
    while(p != NULL) {
        struct ListNode* s = (struct ListNode*)malloc(sizeof(struct ListNode));
        s->next = NULL;
        int sum = p->val + c;
        c = sum / 10;
        s->val = sum % 10;
        s->next = head1->next;
        head1->next = s;
        p = p->next;
    }
    while(q != NULL) {
        struct ListNode* s = (struct ListNode*)malloc(sizeof(struct ListNode));
        s->next = NULL;
        int sum = q->val + c;
        c = sum / 10;
        s->val = sum % 10;
        s->next = head1->next;
        head1->next = s;
        q = q->next;
    }
    if(c) {
        struct ListNode* s = (struct ListNode*)malloc(sizeof(struct ListNode));
        s->next = NULL;
        s->val = c;
        s->next = head1->next;
        head1->next = s;
    }
    return head1->next;
}

运行结果

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