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[LeetCode]2.Add Two Numbers
【題目】
You are given two linked lists representing two non-negative numbers. 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
【題意】
給你兩個鏈表，表示兩個非負整數。數字在鏈表中按反序存儲，例如342在鏈表中為2->4->3。鏈表每一個節點包含一個數字（0-9）。
計算這兩個數字和並以鏈表形式返回。
【分析】
無
【代碼1】
/*********************************
* 日期：2014-01-27
* 作者：SJF0115
* 題目: 2.Add Two Numbers
* 網址：https://oj.leetcode.com/problems/add-two-numbers/
* 結果：AC
* 來源：LeetCode
* 總結：
**********************************/
#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
ListNode *head = (ListNode *)malloc(sizeof(ListNode));
ListNode *pre = head;
ListNode *node = NULL;
//進位
int c = 0,sum;
//加法
while(l1 != NULL && l2 != NULL){
sum = l1->val + l2->val + c;
c = sum / 10;
node = (ListNode *)malloc(sizeof(ListNode));
node->val = sum % 10;
node->next = NULL;
//尾插法
pre->next = node;
pre = node;
l1 = l1->next;
l2 = l2->next;
}
//例如：2->4->3->1 5->6->4
while(l1 != NULL){
sum = l1->val + c;
c = sum / 10;
node = (ListNode *)malloc(sizeof(ListNode));
node->val = sum % 10;
node->next = NULL;
//尾插法
pre->next = node;
pre = node;
l1 = l1->next;
}
//例如：2->4->3 5->6->4->1
while(l2 != NULL){
sum = l2->val + c;
c = sum / 10;
node = (ListNode *)malloc(sizeof(ListNode));
node->val = sum % 10;
node->next = NULL;
//尾插法
pre->next = node;
pre = node;
l2 = l2->next;
}
//最後一位還有進位
if(c > 0){
node = (ListNode *)malloc(sizeof(ListNode));
node->val = c;
node->next = NULL;
//尾插法
pre->next = node;
pre = node;
}
return head->next;
}
};
int main() {
Solution solution;
int A[] = {2,4,7,9};
int B[] = {5,6,4};
ListNode *head = NULL;
ListNode *head1 = (ListNode*)malloc(sizeof(ListNode));
ListNode *head2 = (ListNode*)malloc(sizeof(ListNode));
head1->next = NULL;
head2->next = NULL;
ListNode *node;
ListNode *pre = head1;
for(int i = 0;i < 4;i++){
node = (ListNode*)malloc(sizeof(ListNode));
node->val = A[i];
node->next = NULL;
pre->next = node;
pre = node;
}
pre = head2;
for(int i = 0;i < 3;i++){
node = (ListNode*)malloc(sizeof(ListNode));
node->val = B[i];
node->next = NULL;
pre->next = node;
pre = node;
}
head = solution.addTwoNumbers(head1->next,head2->next);
while(head != NULL){
printf("%d ",head->val);
head = head->next;
}
return 0;
}
【代碼2】
class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
ListNode *head = (ListNode *)malloc(sizeof(ListNode));
ListNode *pre = head;
ListNode *node = NULL;
//進位
int c = 0,sum,val1,val2;
//加法
while(l1 != NULL || l2 != NULL || c != 0){
val1 = (l1 == NULL ? 0 : l1->val);
val2 = (l2 == NULL ? 0 : l2->val);
sum = val1 + val2 + c;
c = sum / 10;
node = (ListNode *)malloc(sizeof(ListNode));
node->val = sum % 10;
node->next = NULL;
//尾插法
pre->next = node;
pre = node;
l1 = (l1 == NULL ? NULL : l1->next);
l2 = (l2 == NULL ? NULL : l2->next);
}
return head->next;
}
};
最後更新：2017-04-03 12:54:49