每天一题LeetCode[第四天]

每天一题LeetCode[第四天]

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Add Two Numbers

Description:

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 Subscribe to see which companies asked this question.

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解题思路：

这题比较简单，思路也是十分容易得出来。 小细节：就是如果用List数据结构，建议用两次for循环来写，逻辑更加清晰，虽然效率相对低了点，但是可读性大大提高。ps:如果有更好的解法，欢迎留言～

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Java代码：

public class Solution { private static final String TAG= Solution.class.getSimpleName(); public static void main(String [] args){ List<Integer> numb1=new ArrayList<>(); List<Integer> numb2=new ArrayList<>(); numb1.add(2); numb1.add(4); numb1.add(3); numb2.add(5); numb2.add(6); numb2.add(4); List<Integer> result=addTwoNumbs(numb1,numb2); System.out.println(result.toString()); } public static List<Integer> addTwoNumbs(List<Integer> numb1, List<Integer> numb2){ if(null==numb1 || null==numb2){ return Collections.emptyList(); } List<Integer> lNumbs,sNumbs; if(numb1.size()<numb2.size()){ lNumbs=numb2; sNumbs=numb1; }else{ lNumbs=numb1; sNumbs=numb2; } //第一次循环 不做判断，单纯全部加起来 for(int i=0; i<sNumbs.size();i++){ lNumbs.set(i,lNumbs.get(i)+sNumbs.get(i)); } //第二次循环 判断数值是否大于10 大于则进位 boolean bit=false; for(int i=0;i<lNumbs.size();i++){ int num=lNumbs.get(i); if(bit){ if(i==lNumbs.size()-1 && num==9){ lNumbs.add(1); }else{ num++; } } if(num>=10){ bit=true; }else{ bit=false; } lNumbs.set(i,num%10); } return lNumbs ; } }

提高代码质量就是：每天积累精美的思路，优质的细节的过程。