Problem: 2. 兩數(shù)相加

思路

主要是一一相加和逆序的方式存儲
先說逆序儲存，看下圖
我們先聲明出指針p和指針q，還有指針head（主要用于return上而已），然后進(jìn)行一系列操作，之后，p = q，之后的操作就是對q進(jìn)行，至于p，只做一個動作，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！好說，將對應(yīng)的兩個數(shù)加起來，然后檢查是否大于9，如果是，則要保留個位，并進(jìn)一位（也就是將十位數(shù)的部分放入下一個相加的過程中），以指針p的部分為例——指針q的部分也一樣。

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與l2終會到達(dá)null，所以，對于剩余的部分也只是解決前面的“進(jìn)一位”遺留的問題而已.

    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，倘若還存在進(jìn)一位，就得要再建一個節(jié)點(diǎn)。

    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;
    }

解題方法

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

其實，當(dāng)時在解決這道題的時候，碰到過這樣的問題

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>

后來，在我看了AuthurLEE的文章之后，我才知道原來是忘記在初始化的時候，讓指針指向NULL了（包括head指針，還有結(jié)構(gòu)體里面的next）文章來源地址http://www.zghlxwxcb.cn/news/detail-835680.html

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