leetcode题解日练--2016.8.30

来源：互联网发布：大学算法教程编辑：程序博客网时间：2024/05/04 04:16

###不给自己任何借口

今日题目：

1、字符串乘法；

2、两个链表加法

今日摘录：

年轻吗不要紧
听过几首歌爱过几个人就老了

——独木舟《深海里的星星Ⅱ》

43. Multiply Strings | Difficulty: Medium

Given two numbers represented as strings, return multiplication of the numbers as a string.

Note:
The numbers can be arbitrarily large and are non-negative.
Converting the input string to integer is NOT allowed.
You should NOT use internal library such as BigInteger.
tag：数学|字符串
题意：字符串之间的乘法，不允许转换为整型。

思路：
1、题目明确说了不能转整型，可以考虑https://discuss.leetcode.com/topic/30508/easiest-java-solution-with-graph-explanation/2的解法

class Solution {public:    string multiply(string num1, string num2) {        int m = num1.size(),n=num2.size();        int *pos = new int[m+n]();        for(int i=m-1;i>=0;i--)        {            for(int j=n-1;j>=0;j--)            {                int posHigh = i+j,posLow = i+j+1;                int mul = (num1[i]-'0')*(num2[j]-'0')+pos[posLow];                pos[posHigh] += mul/10;                pos[posLow] = (mul)%10;            }        }        string res="";        for(int i=0;i<m+n;i++)            res =res+char(pos[i]-0+'0');        while(res[0]=='0')        {            res = res.substr(1,res.size()-1);        }        return res.size()==0?"0":res;    }};

结果：8ms

2. Add Two Numbers | Difficulty: Medium

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

tag：链表|数学

题意：两个链表都是逆序存储数位，求它们的加法
思路：
1、两个链表的长度可能相等也可能不相等，分两种情况讨论：
链表长度相等的时候，直接依次访问链表中元素，一定会同时访问完两个链表，最后判断下有没有进位来决定是否还需要加一个1.
链表长度不相等的时候，一定最后还有一个链表没有访问完毕，继续对该链表的元素进行访问，最后判断下有没有进位来决定是否还需要加一个1.

/** * 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) {        if(l1==NULL || l2==NULL) return l1==NULL?l2:l1;        ListNode *res = new ListNode(INT_MIN);        ListNode*cur = res;        int c=0;        while(l1 && l2)        {            cur->next = l2;            int digit = l1->val+l2->val+c;            l2->val = digit%10;            c = digit/10;            cur = cur->next;            l1 = l1->next;            l2 = l2->next;        }        if(l1||l2)          {            if(l1)  cur->next = l1;            cur = cur->next;            while(c&&cur)            {                int digit = (cur->val+c);                cur->val = digit%10;                c = digit/10;                if(cur->next)  cur = cur->next;                else    break;            }        }        if(c)  {cur->next=new ListNode(1);}        return res->next;    }};

结果：48ms

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