LeetCode题解--2-Add Two Numbers
2016-03-05 16:32
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题目地址 https://leetcode.com/problems/add-two-numbers/
题解 https://github.com/gatieme/LeetCode/tree/master/2-AddTwoNumbers
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
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这道题其实是大数相加的处理,没什么难度,但需要注意以下几点:
因为存储是反过来的,即数字342存成2->4->3,所以要注意进位是向后的;
链表l1和l2长度可能不同,因此要注意处理某个链表剩余的高位;
最高位如果有进位,那么需要新增一个结点存储最高进位
每一位都是0~9的数字,不会超过一个位
这样的话,我们两个整数可以直接求和,而不用考虑各位的进位问题
最后再进行处理,将和再转换为链表
题解 https://github.com/gatieme/LeetCode/tree/master/2-AddTwoNumbers
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
Subscribe to see which companies asked this question
这道题其实是大数相加的处理,没什么难度,但需要注意以下几点:
因为存储是反过来的,即数字342存成2->4->3,所以要注意进位是向后的;
链表l1和l2长度可能不同,因此要注意处理某个链表剩余的高位;
最高位如果有进位,那么需要新增一个结点存储最高进位
每一位都是0~9的数字,不会超过一个位
思路
链表单独进位处理
C++
思路非常简单,利用两个指针分别遍历两个链表,并且用一个变量表示是否有进位。某个链表遍历结束之后再将另一个链表连接在结果链表之后即可,若最后有进位需要添加一位。[code]#include <cstdlib> #include <iostream> #include <cmath> using namespace std; #define DEBUG #ifdef DEBUG class ListNode { public : ListNode( ) :val(0), next(NULL) { } ListNode(int x) :val(x), next(NULL) { } int val; ListNode *next; }; #endif /** * 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) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. if((l1 == NULL) && (l2 == NULL)) { return NULL; } int sum = 0; int carry = 0; // 进位 int current = 0; ListNode *p = new ListNode(0); ListNode *start = p; sum = l1->val + l2->val + carry; carry = sum / 10; p->val = sum % 10; l1 = l1->next; l2 = l2->next; while(l1 != NULL && l2 != NULL) { sum = l1->val + l2->val + carry; carry = sum / 10; p->next = new ListNode((int)sum % 10); p = p->next; l1 = l1->next; l2 = l2->next; } // 处理多出的位,因为两个链表可能不一样长度 if(l1 != NULL) { while(l1 != NULL) { sum = l1->val + carry; carry = sum / 10; p->next = new ListNode((int)sum % 10); p = p->next; l1 = l1->next; } } else if(l2 != NULL) { while(l2 != NULL) { sum = l2->val + carry; carry = sum / 10; p->next = new ListNode((int)sum % 10); p = p->next; l2 = l2->next; } } if(carry != 0) { p->next = new ListNode((int)carry); p = p->next; } return start; } }; #ifdef DEBUG int main( int argc, char *argv[] ) { ListNode l1[5], l2[5]; ListNode *pl1 = &l1[0], *pl2 = &l2[0]; // Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) // Output: 7 -> 0 -> 8 l1[0].val = 2;l1[0].next = &l1[1]; l1[1].val = 4;l1[1].next = &l1[2]; l1[2].val = 7;l1[2].next = NULL; l2[0].val = 5;l2[0].next = &l2[1]; l2[1].val = 6;l2[1].next = &l2[2]; l2[2].val = 7;l2[2].next = NULL; Solution solu = Solution(); ListNode *currNode = solu.addTwoNumbers(pl1, pl2); std::cout <<currNode->val; while(currNode->next != NULL) { currNode = currNode->next; std::cout <<" -> " <<currNode->val; } return EXIT_SUCCESS; } #endif
Python
[code]#!/usr/bin/env python #coding=utf-8 # class ListNode: def __init__(self, x): self.val = x self.next = None class Solution: # @return a ListNode def addTwoNumbers(self, l1, l2): ret = ListNode(0) cur = ret sum = 0 while True: if l1 != None: sum += l1.val l1 = l1.next if l2 != None: sum += l2.val l2 = l2.next cur.val = sum % 10 sum /= 10 if l1 != None or l2 != None or sum != 0: cur.next = ListNode(0) cur = cur.next else: break return ret # debug s = Solution()
换算成整数求和
将两个链表直接换算成两个整数这样的话,我们两个整数可以直接求和,而不用考虑各位的进位问题
最后再进行处理,将和再转换为链表
[code]#include <cstdlib> #include <iostream> #include <cmath> using namespace std; #define DEBUG #ifdef DEBUG class ListNode { public : ListNode( ) :val(0), next(NULL) { } ListNode(int x) :val(x), next(NULL) { } int val; ListNode *next; }; #endif /** * 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) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. if((l1 == NULL) && (l2==NULL)) { return NULL; } unsigned long long sum1 = 0; unsigned long long sum2 = 0; unsigned long long sum3 = 0; // 位数 int count1 = 0; int count2 = 0; // 2 -> 4 -> 3 while(l1 != NULL) { sum1 += pow(10, count1) * (l1->val); count1++; l1 = l1->next; } while(l2 != NULL) { sum2 += pow(10, count2) * (l2->val); count2++; l2=l2->next; } sum3 = sum1 + sum2; #ifdef DEBUG std::cout <<sum1 <<" + " <<sum2 <<" = " <<sum3 <<std::endl; #endif unsigned long tmp_digit = sum3 % 10; sum3 = sum3 / 10; ListNode* p = new ListNode((int)tmp_digit); ListNode* start = p; while(sum3 > 0) { tmp_digit = sum3 % 10; sum3 /= 10; p->next = new ListNode((int)tmp_digit); p = p->next; } p->next = NULL; //printList(start); return start; } }; #ifdef DEBUG int main( int argc, char *argv[] ) { ListNode l1[3], l2[3]; ListNode *pl1 = &l1[0], *pl2 = &l2[0]; // Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) // Output: 7 -> 0 -> 8 l1[0].val = 2;l1[0].next = &l1[1]; l1[1].val = 4;l1[1].next = &l1[2]; l1[2].val = 3;l1[2].next = NULL; l2[0].val = 5;l2[0].next = &l2[1]; l2[1].val = 6;l2[1].next = &l2[2]; l2[2].val = 4;l2[2].next = NULL; Solution solu = Solution(); ListNode *currNode = solu.addTwoNumbers(pl1, pl2); std::cout <<currNode->val; while(currNode->next != NULL) { currNode = currNode->next; std::cout <<" -> " <<currNode->val; } return EXIT_SUCCESS; } #endif
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