[LeetCode]Add Two Numbers

Posted 2020-09-08 乘风有时

链表的数据对应位置相加，之和组成新的链表，已经说过是非负的值。

题目不难，注意链表的操作。

本人觉得平时练习有时间，即便题目说链表不为空，我们也应该判断一下，养成良好的习惯很重要。

说一下注意点：

1.主要考虑进位

2.注意可能链表不一样长

3.注意不要忘记了最高位的进位

这里用的是尾插法，可以考虑用头插法试试。

  1 /***************************************
  2 You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
  3 You may assume the two numbers do not contain any leading zero, except the number 0 itself.
  4 
  5 Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
  6 Output: 7 -> 0 -> 8
  7 ***************************************/
  8 
  9 /**
 10  * Definition for singly-linked list.
 11  * struct ListNode {
 12  *     int val;
 13  *     struct ListNode *next;
 14  * };
 15  */
 16 #include<stdio.h>
 17 
 18 struct ListNode {
 19     int val;
 20     struct ListNode *next;
 21 };
 22  
 23 struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
 24     if(l1 == NULL)return l2;
 25     if(l2 == NULL)return l1;
 26     int j = 0;//record carry value记录进位值
 27     struct ListNode* ln = (struct ListNode *)malloc(sizeof(struct ListNode));
 28     struct ListNode* p = ln;
 29     struct ListNode* p1 = l1;
 30     struct ListNode* p2 = l2;
 31     int i = p1->val + p2->val + j;
 32     p->val = i%10;
 33     j = i/10;
 34     p1 = p1->next;
 35     p2 = p2->next;
 36     while(p1 != NULL && p2 != NULL){//尾插法
 37         struct ListNode *q = (struct ListNode *)malloc(sizeof(struct ListNode));
 38         p->next = q;
 39         p=q;
 40         i = p1->val + p2->val + j;
 41         p->val = i%10;
 42         j = i/10;
 43         p1 = p1->next;
 44         p2 = p2->next;
 45     }
 46 
 47     while(p1 != NULL){//l1剩余未转换完的数字
 48         struct ListNode *q = (struct ListNode *)malloc(sizeof(struct ListNode));
 49         p->next = q;
 50         p=q;
 51         i = p1->val + j;
 52         p->val = i%10;
 53         j = i/10;
 54         p1 = p1->next;
 55     }
 56     while(p2 != NULL){//l2剩余未转换完的数字
 57         struct ListNode *q = (struct ListNode *)malloc(sizeof(struct ListNode));
 58         p->next = q;
 59         p=q;
 60         i = p2->val + j;
 61         p->val = i%10;
 62         j = i/10;
 63         p2 = p2->next;
 64     }
 65     if(j > 0){//最高位进位
 66         struct ListNode *q = (struct ListNode *)malloc(sizeof(struct ListNode));
 67         p->next = q;
 68         p=q;
 69         p->val = j;
 70     }
 71     p->next = NULL;
 72     return ln;
 73 }
 74 
 75 int main(){
 76     struct ListNode* l1 = (struct ListNode *)malloc(sizeof(struct ListNode));
 77     struct ListNode* l2 = (struct ListNode *)malloc(sizeof(struct ListNode));
 78     struct ListNode* p1 = l1;
 79     struct ListNode* p2 = l2;
 80     int i = 3455;
 81     int j = 16789;
 82     l1->val = i%10;
 83     l2->val = j%10;
 84     i = i/10;
 85     j = j/10;
 86     printf("%d:%d\n",p1->val,p2->val);
 87     while (i > 0 && j > 0)
 88     {
 89         struct ListNode* q1 = (struct ListNode *)malloc(sizeof(struct ListNode));
 90         q1->val = i%10;
 91         i = i/10;
 92         p1->next = q1;
 93         p1 = q1;
 94         struct ListNode* q2 = (struct ListNode *)malloc(sizeof(struct ListNode));
 95         q2->val = j%10;
 96         j = j/10;
 97         p2->next = q2;
 98         p2 = q2;
 99         printf("%d:%d\n",p1->val,p2->val);
100     }
101     while(i > 0){
102         struct ListNode* q1 = (struct ListNode *)malloc(sizeof(struct ListNode));
103         q1->val = i%10;
104         i = i/10;
105         p1->next = q1;
106         p1 = q1;
107     }
108     while(j > 0){
109         struct ListNode* q2 = (struct ListNode *)malloc(sizeof(struct ListNode));
110         q2->val = i%10;
111         j = j/10;
112         p2->next = q2;
113         p2 = q2;
114     }
115     p1->next = NULL;
116     p2->next = NULL;
117     
118     struct ListNode* ap = addTwoNumbers(l1,l2);
119     struct ListNode* aq = ap;
120     p1 = l1;
121     p2 = l2;
122     while(aq != NULL){
123         printf("%d -> ",aq->val);
124         aq = aq->next;
125     }
126     
127     while(ap != NULL){
128         aq = ap;
129         ap = ap->next;
130         free(aq);
131     }
132     while(l1 != NULL){
133         p1 = l1;
134         l1 = l1->next;
135         free(p1);
136     }
137     while(l2 != NULL){
138         p2 = l2;
139         l2 = l2->next;
140         free(p2);
141     }
142     
143     return 0;
144 }