LeetCode 2. Add Two Numbers

2019-11-06 07:36:35

字体：大中小

来源：转载

供稿：网友

2. Add Two Numbers

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

题目描述：给出两个数字链表，求出这两个数字的和并且以类似的链表返回。这里的链表里面的数字顺序是倒序的，也就是说第一个是个位，第二个是十位，第三个是百位…… 题目给出了这个链表的结构如下：

//Definition for singly-linked list.struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {}};

解题思路：我的做法是直接将两个链表的数字逐位相加，并且保存两个数字相加后的进位情况，如果相加后的数是一个两位数，那么要将这个数对10取模保留个位数的值，然后把进位标志设为1，在下一位相加的时候把这个进位也加上去。

代码：

#include <iostream>#include <string>using namespace std;//Definition for singly-linked list.struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {}};class Solution {public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { ListNode* result = NULL; ListNode* temp = NULL; int carry = 0; int current = 0; while (l1 != NULL || l2 != NULL) { current = 0; if (l1 != NULL) { current += l1->val; l1 = l1->next; } if (l2 != NULL) { current += l2->val; l2 = l2->next; } current += carry; carry = current / 10; current %= 10; if (result == NULL) { result = new ListNode(current); temp = result; } else { result->next = new ListNode(current); result = result->next; } if (l1 == NULL && l2 == NULL && carry != 0) { result->next = new ListNode(carry); } } return temp; }};void PrintList(ListNode* l) { if (l == NULL) return; ListNode* currentNode = l; while (currentNode != NULL) { if (currentNode != l) { cout << " -> "; } cout << currentNode->val; currentNode = currentNode->next; }}ListNode* Convert2ListNode(string val) { if (val == "") return NULL; ListNode* l = new ListNode(val[val.length() - 1] - '0'); ListNode* start = l; std::string::reverse_iterator r_it = val.rbegin() + 1; for (; r_it != val.rend(); r_it++) { l->next = new ListNode(*r_it - '0'); l = l->next; } return start;}int main() { Solution sln; ListNode* l1 = Convert2ListNode("5"); ListNode* l2 = Convert2ListNode("5"); PrintList(l1); cout << endl; PrintList(l2); cout << endl; ListNode* l3 = sln.addTwoNumbers(l1, l2); PrintList(l3); return 0;}

上一篇：使用crypoto++进行加密解密签名和认证心得

下一篇：Leetcode 61. Rotate List （反转链表）

学习交流

索泰发布一款GTX 1070 Mini迷你版本:小机

索泰发布一款GTX 1070 Mini迷你版本:小机箱大爱...

热门图片

猜你喜欢的新闻

猜你喜欢的关注