LeetCode | 两数相加 C语言

Problem: 2. 两数相加

文章目录

• 思路
• 解题方法
• Code
• 一些感想

思路

主要是一一相加和逆序的方式存储

先说逆序储存，看下图

我们先声明出指针p和指针q，还有指针head（主要用于return上而已），然后进行一系列操作，之后，p = q，之后的操作就是对q进行，至于p，只做一个动作，p = q

	int flag = 0;
	struct ListNode *p = NULL, *q = NULL, *head = NULL;
    while(l1 != NULL && l2 != NULL) {
        if(flag == 0) {
            flag = 1;
            p = (struct ListNode*)malloc(sizeof(struct ListNode));
            p->next = NULL;
            head = p;
        } else {
            q = (struct ListNode*)malloc(sizeof(struct ListNode));
            q->next = NULL;
            p->next = q;
            p = q;
        }
        l1 = l1->next;
        l2 = l2->next;
    }

至于相加嘛！好说，将对应的两个数加起来，然后检查是否大于9，如果是，则要保留个位，并进一位（也就是将十位数的部分放入下一个相加的过程中），以指针p的部分为例------指针q的部分也一样。

p->val = l1->val + l2->val;
if(p->val > 9) {
	last = p->val / 10;
	p->val %= 10;
} else last = 0;

但是，要知道，l1与l2终会到达null，所以，对于剩余的部分也只是解决前面的"进一位"遗留的问题而已.

    while(l1 != NULL) {
        q = (struct ListNode*)malloc(sizeof(struct ListNode));
        q->val = l1->val + last;
        if(q->val > 9) {
            last = q->val / 10;
            q->val %= 10;
        } else last = 0;
        q->next =NULL;
        p->next = q;
        p = q;
        l1 = l1->next;
    }

    while(l2 != NULL) {
        q = (struct ListNode*)malloc(sizeof(struct ListNode));
        q->val = l2->val + last;
        if(q->val > 9) {
            last = q->val / 10;
            q->val %= 10;
        } else last = 0;
        q->next =NULL;
        p->next = q;
        p = q;
        l2 = l2->next;
    }

到最后，倘若还存在进一位，就得要再建一个节点。

    if(last > 0) {
        q = (struct ListNode*)malloc(sizeof(struct ListNode));
        q->val = last;
        if(q->val > 9) {
            last = q->val / 10;
            q->val %= 10;
        } else last = 0;
        q->next =NULL;
        p->next = q;
        p = q;
    }

解题方法

由思路可知

Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode *p = NULL, *q = NULL, *head = NULL;
    int flag = 0, last;
    while(l1 != NULL && l2 != NULL) {
        if(flag == 0) {
            flag = 1;
            p = (struct ListNode*)malloc(sizeof(struct ListNode));
            p->val = l1->val + l2->val;
            if(p->val > 9) {
                last = p->val / 10;
                p->val %= 10;
            } else last = 0;
            p->next = NULL;
            head = p;
        } else {
            q = (struct ListNode*)malloc(sizeof(struct ListNode));
            q->val = l1->val + l2->val + last;
            if(q->val > 9) {
                last = q->val / 10;
                q->val %= 10;
            } else last = 0;
            q->next =NULL;
            p->next = q;
            p = q;
        }
        l1 = l1->next;
        l2 = l2->next;
    }
    while(l1 != NULL) {
        q = (struct ListNode*)malloc(sizeof(struct ListNode));
        q->val = l1->val + last;
        if(q->val > 9) {
            last = q->val / 10;
            q->val %= 10;
        } else last = 0;
        q->next =NULL;
        p->next = q;
        p = q;
        l1 = l1->next;
    }
    while(l2 != NULL) {
        q = (struct ListNode*)malloc(sizeof(struct ListNode));
        q->val = l2->val + last;
        if(q->val > 9) {
            last = q->val / 10;
            q->val %= 10;
        } else last = 0;
        q->next =NULL;
        p->next = q;
        p = q;
        l2 = l2->next;
    }
    if(last > 0) {
        q = (struct ListNode*)malloc(sizeof(struct ListNode));
        q->val = last;
        if(q->val > 9) {
            last = q->val / 10;
            q->val %= 10;
        } else last = 0;
        q->next =NULL;
        p->next = q;
        p = q;
    }
    return head;
}

一些感想

其实，当时在解决这道题的时候，碰到过这样的问题

Line 70: Char 15: runtime error: member access within misaligned address 0xbebebebebebebebe for type 'struct ListNode', which requires 8 byte alignment [ListNode.c]0xbebebebebebebebe: note: pointer points here<memory cannot be printed>

后来，在我看了AuthurLEE的文章之后，我才知道原来是忘记在初始化的时候，让指针指向NULL了（包括head指针，还有结构体里面的next）