描述
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
分析：342+465=807——>708
Add.h
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#pragma once
#include <iostream>
using namespace std;
typedef struct ListNode{
int _var;
ListNode *_next;
ListNode(int var)
:_var(var)
, _next(NULL)
{}
}node, *node_p;
class Solution
{
public:
node_p add_two_number(ListNode *l1, ListNode *l2)
{
node_p NewHead = NULL;
node_p pa = l1->_next;
node_p pb = l2->_next;
int mod = 0;
while (pa != NULL || pb != NULL){
int a = 0;
int b = 0;
if (pa != NULL)
a = pa->_var;
if (pb != NULL)
b = pb->_var;
int ret = (a + b + mod) % 10;
mod = (a + b + mod) / 10;
node_p tmp = new node(ret);
tmp->_next = NewHead;
NewHead = tmp;
if (pa != NULL)
pa = pa->_next;
if (pb != NULL)
pb = pb->_next;
}
if (mod > 0){
node_p tmp = new node(mod);
tmp->_next = NewHead;
NewHead = tmp;
}
return NewHead;
}
};Add.cpp
#include "Add_List.h"
#include <stdlib.h>
using namespace std;
int main()
{
Solution s1;
node_p l1 = new node(-1);
node_p pa = l1;
pa->_next = new node(4);
pa = pa->_next;
pa->_next = new node(6);
pa = pa->_next;
pa->_next = new node(7);
pa = pa->_next;
pa->_next = new node(8);
node_p l2 = new node(-1);
node_p pb = l2;
pb->_next = new node(1);
pb = pb->_next;
pb->_next = new node(2);
pb = pb->_next;
pb->_next = new node(5);
pb = pb->_next;
pb->_next = new node(6);
pb = pb->_next;
pb->_next = new node(3);
node_p ret=s1.add_two_number(l1,l2);
system("pause");
return 0;
}调试查看结果：
// LeetCode, Add Two Numbers
// 时间复杂度 O(m+n)，空间复杂度 O(1)
class Solution{
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
ListNode dummy(-1); // 头节点
int carry = 0;
ListNode *prev = &dummy;
for (ListNode *pa = l1, *pb = l2;
pa != nullptr || pb != nullptr;
pa = pa == nullptr ? nullptr : pa->next,
pb = pb == nullptr ? nullptr : pb->next,
prev = prev->next) {
const int ai = pa == nullptr ? 0 : pa->val;
const int bi = pb == nullptr ? 0 : pb->val;
const int value = (ai + bi + carry) % 10;
carry = (ai + bi + carry) / 10;
prev->next = new ListNode(value); // 尾插法
}
if (carry > 0)
prev->next = new ListNode(carry);
return dummy.next;
}
};《完》
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