Add Two Numbers --leetcode
2017-03-16 22:05
357 查看
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.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
函数接口:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
}
我的函数:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
char f=0;
unsigned int i=0,j=0;
struct ListNode *L,*L_long,*L_short;
L=l1;//统计l1数据长度i
while(L!=NULL){
i++;
L=L->next;
}
L=l2;//统计l2数据长度j
while(L!=NULL){
j++;
L=L->next;
}
if(i>j){
L_short=l2;
L_long=l1;
}
else{
L_short=l1;
L_long=l2;
}
L=L_long;
do{
if(f==1){
L_long->val+=L_short->val;
L_long->val+=1;
// printf("%d ",L_long->val);
f=0;
}
else{
L_long->val+=L_short->val;
// printf("%d ",L_long->val);
}
if(L_long->val>=10){
L_long->val%=10;
f=1;
}
L_short=L_short->next;
if(L_short!=NULL){
L_long=L_long->next;
}
}while(L_short!=NULL);
aa: if(f==1&&L_long->next==NULL){ //判断L_long是否为链表的最后一个节点,是最后一个节点,进位--添加节点。
f=0;
struct ListNode *s;
s=(struct ListNode *)malloc(sizeof(struct ListNode));
L_long->next=s;
s->val=1;
s->next=NULL;
}
if(f==1&&L_long->next!=NULL){ //判断L_long是否为链表的最后一个节点,不是最后一个节点。
f=0;
L_long=L_long->next;
L_long->val+=1;
if(L_long->val==10){ //判断是否进位,进位。
f=1;
L_long->val=0;
goto aa;
}
}
return L;
}
自己写的测试函数(模拟这个题目的测试环境):
#include<stdlib.h>
#include<stdio.h>
struct ListNode {
int val;
struct ListNode *next;
};
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
char f=0;
unsigned int i=0,j=0;
struct ListNode *L,*L_long,*L_short;
L=l1;//统计l1数据长度i
while(L!=NULL){
i++;
L=L->next;
}
L=l2;//统计l2数据长度j
while(L!=NULL){
j++;
L=L->next;
}
if(i>j){
L_short=l2;
L_long=l1;
}
else{
L_short=l1;
L_long=l2;
}
L=L_long;
do{
if(f==1){
L_long->val+=L_short->val;
L_long->val+=1;
// printf("%d ",L_long->val);
f=0;
}
else{
L_long->val+=L_short->val;
// printf("%d ",L_long->val);
}
if(L_long->val>=10){
L_long->val%=10;
f=1;
}
L_short=L_short->next;
if(L_short!=NULL){
L_long=L_long->next;
}
}while(L_short!=NULL);
aa: if(f==1&&L_long->next==NULL){ //判断L_long是否为链表的最后一个节点,是最后一个节点,进位--添加节点。
f=0;
struct ListNode *s;
s=(struct ListNode *)malloc(sizeof(struct ListNode));
L_long->next=s;
s->val=1;
s->next=NULL;
}
if(f==1&&L_long->next!=NULL){ //判断L_long是否为链表的最后一个节点,不是最后一个节点。
f=0;
L_long=L_long->next;
L_long->val+=1;
if(L_long->val==10){ //判断是否进位,进位。
f=1;
L_long->val=0;
goto aa;
}
}
return L;
}
void main(){
struct ListNode *l1,*l2;
struct ListNode *L;
struct ListNode *p;
int i,j;
l1=(struct ListNode *)malloc(sizeof(struct ListNode));
l1->next=NULL;
printf("Please input the length of the l1:");
scanf("%d",&i);
for(;i>0;i--){
p=(struct ListNode *)malloc(sizeof(struct ListNode));
scanf("%d",&p->val);
p->next=l1->next;l1->next=p;
}
l1=l1->next;//表头非空
l2=(struct ListNode *)malloc(sizeof(struct ListNode));
l2->next=NULL;
printf("Please input the length of the l2:");
scanf("%d",&j);
for(;j>0;j--){
p=(struct ListNode *)malloc(sizeof(struct ListNode));
scanf("%d",&p->val);
p->next=l2->next;l2->next=p;
}
l2=l2->next;//表头非空
L=addTwoNumbers(l1,l2);
printf("\nThe resault is:");
while(L->next!=NULL){
printf("%d->",L->val);
L=L->next;
}
printf("%d\n",L->val);
return;
}
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
函数接口:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
}
我的函数:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
char f=0;
unsigned int i=0,j=0;
struct ListNode *L,*L_long,*L_short;
L=l1;//统计l1数据长度i
while(L!=NULL){
i++;
L=L->next;
}
L=l2;//统计l2数据长度j
while(L!=NULL){
j++;
L=L->next;
}
if(i>j){
L_short=l2;
L_long=l1;
}
else{
L_short=l1;
L_long=l2;
}
L=L_long;
do{
if(f==1){
L_long->val+=L_short->val;
L_long->val+=1;
// printf("%d ",L_long->val);
f=0;
}
else{
L_long->val+=L_short->val;
// printf("%d ",L_long->val);
}
if(L_long->val>=10){
L_long->val%=10;
f=1;
}
L_short=L_short->next;
if(L_short!=NULL){
L_long=L_long->next;
}
}while(L_short!=NULL);
aa: if(f==1&&L_long->next==NULL){ //判断L_long是否为链表的最后一个节点,是最后一个节点,进位--添加节点。
f=0;
struct ListNode *s;
s=(struct ListNode *)malloc(sizeof(struct ListNode));
L_long->next=s;
s->val=1;
s->next=NULL;
}
if(f==1&&L_long->next!=NULL){ //判断L_long是否为链表的最后一个节点,不是最后一个节点。
f=0;
L_long=L_long->next;
L_long->val+=1;
if(L_long->val==10){ //判断是否进位,进位。
f=1;
L_long->val=0;
goto aa;
}
}
return L;
}
自己写的测试函数(模拟这个题目的测试环境):
#include<stdlib.h>
#include<stdio.h>
struct ListNode {
int val;
struct ListNode *next;
};
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
char f=0;
unsigned int i=0,j=0;
struct ListNode *L,*L_long,*L_short;
L=l1;//统计l1数据长度i
while(L!=NULL){
i++;
L=L->next;
}
L=l2;//统计l2数据长度j
while(L!=NULL){
j++;
L=L->next;
}
if(i>j){
L_short=l2;
L_long=l1;
}
else{
L_short=l1;
L_long=l2;
}
L=L_long;
do{
if(f==1){
L_long->val+=L_short->val;
L_long->val+=1;
// printf("%d ",L_long->val);
f=0;
}
else{
L_long->val+=L_short->val;
// printf("%d ",L_long->val);
}
if(L_long->val>=10){
L_long->val%=10;
f=1;
}
L_short=L_short->next;
if(L_short!=NULL){
L_long=L_long->next;
}
}while(L_short!=NULL);
aa: if(f==1&&L_long->next==NULL){ //判断L_long是否为链表的最后一个节点,是最后一个节点,进位--添加节点。
f=0;
struct ListNode *s;
s=(struct ListNode *)malloc(sizeof(struct ListNode));
L_long->next=s;
s->val=1;
s->next=NULL;
}
if(f==1&&L_long->next!=NULL){ //判断L_long是否为链表的最后一个节点,不是最后一个节点。
f=0;
L_long=L_long->next;
L_long->val+=1;
if(L_long->val==10){ //判断是否进位,进位。
f=1;
L_long->val=0;
goto aa;
}
}
return L;
}
void main(){
struct ListNode *l1,*l2;
struct ListNode *L;
struct ListNode *p;
int i,j;
l1=(struct ListNode *)malloc(sizeof(struct ListNode));
l1->next=NULL;
printf("Please input the length of the l1:");
scanf("%d",&i);
for(;i>0;i--){
p=(struct ListNode *)malloc(sizeof(struct ListNode));
scanf("%d",&p->val);
p->next=l1->next;l1->next=p;
}
l1=l1->next;//表头非空
l2=(struct ListNode *)malloc(sizeof(struct ListNode));
l2->next=NULL;
printf("Please input the length of the l2:");
scanf("%d",&j);
for(;j>0;j--){
p=(struct ListNode *)malloc(sizeof(struct ListNode));
scanf("%d",&p->val);
p->next=l2->next;l2->next=p;
}
l2=l2->next;//表头非空
L=addTwoNumbers(l1,l2);
printf("\nThe resault is:");
while(L->next!=NULL){
printf("%d->",L->val);
L=L->next;
}
printf("%d\n",L->val);
return;
}
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