leetcode题目-求两相加

发表于 2024/03/27 10:24:16 [计算机基础] 浏览次数：116

2.两数相加

给出两个非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照逆序的方式存储的，并且它们的每个节点只能存储一位数字。

如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。

您可以假设除了数字 0 之外，这两个数都不会以 0开头。

示例：

输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807

来源：力扣（LeetCode）

链接：https://leetcode-cn.com/problems/add-two-numbers

解题：

#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int val;
    struct ListNode *next;
};

static void showNode(struct ListNode* head){
    while (head){
        printf("%d->", head->val);
        head = head->next;
    }
}
/**
 * 两数相加
 * @param l1
 * @param l2
 * @return
 */

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
    if(l1 == NULL){
        return l2;
    }
    if(l2 == NULL){
        return l1;
    }
    struct ListNode* temp1 = l1;
    struct ListNode* temp2 = l2;
    // 是否进位
    int isCarry  = 0;
    struct ListNode* head = NULL;
    struct ListNode* current = NULL;
    struct ListNode* currentPre = NULL;
    int num = 0;
    while (temp1 && temp2){
        num = temp1->val + temp2->val + isCarry;
        current = (struct ListNode*)malloc(sizeof(struct ListNode));
        if(currentPre!= NULL){
            currentPre->next = current;
        }
        if(head == NULL){
            head = current;
        }
        if(num >= 10){
            num = num-10;
            isCarry = 1;
        }else {
            isCarry = 0;
        }
        current->val = num;
        current->next = NULL;
        temp1 = temp1->next;
        temp2 = temp2->next;
        currentPre = current;
    }
    while (temp1){
        num = temp1->val  + isCarry;
        current = (struct ListNode*)malloc(sizeof(struct ListNode));
        currentPre->next = current;
        if(num >= 10){
            num = num-10;
            isCarry = 1;
        }else {
            isCarry = 0;
        }
        current->val = num;
        current->next = NULL;
        temp1 = temp1->next;
        currentPre = current;
    }
    while (temp2){
        num = temp2->val  + isCarry;
        current = (struct ListNode*)malloc(sizeof(struct ListNode));
        currentPre->next = current;
        if(num >= 10){
            num = num-10;
            isCarry = 1;
        }else {
            isCarry = 0;
        }
        current->val = num;
        current->next = NULL;
        temp2 = temp2->next;
        currentPre = current;
    }
    if(isCarry){
        current = (struct ListNode*)malloc(sizeof(struct ListNode));
        currentPre->next = current;
        current->val = 1;
        current->next = NULL;
    }

    return head;

}
int main(){
    struct ListNode* l1 = (struct ListNode*)malloc(sizeof(struct ListNode));
    struct ListNode* l2 = (struct ListNode*)malloc(sizeof(struct ListNode));
    l1->val = 1;
    l2->val = 2;
    struct ListNode* temp1 = l1;
    struct ListNode* temp2 = l2;
    for(int i = 0; i < 3; i++){
        temp1->next = (struct ListNode*)malloc(sizeof(struct ListNode));
        temp2->next = (struct ListNode*)malloc(sizeof(struct ListNode));
        temp1->next->val = i+6;
        temp1->next->next = NULL;
        temp2->next->val = i+6;
        temp2->next->next = NULL;
        temp1 = temp1->next;
        temp2 = temp2->next;
    }

    showNode(addTwoNumbers(l1, l2));
    return 0;
}