• 【LeetCode】66 & 67- Plus One & Add Binary


    66 - Plus One

    Given a non-negative number represented as an array of digits, plus one to the number.

    The digits are stored such that the most significant digit is at the head of the list.

    Solution 1 : 十进制加法

     1 class Solution {
     2 public:
     3     vector<int> plusOne(vector<int> &digits) {
     4         int carry = 1;
     5         for(int i = digits.size()-1; i >= 0; i --)
     6         {
     7             int sum = digits[i]+carry;
     8             carry = sum / 10;
     9             digits[i] = sum % 10;
    10             if(carry == 0)
    11                 break;
    12         }
    13         if(carry == 1)
    14             digits.insert(digits.begin(), 1);
    15         return digits;
    16     }
    17 };

    Solution 2 :

     1 class Solution {
     2 public:
     3     vector<int> plusOne(vector<int> &digits) {
     4         for(int i = digits.size()-1; i >= 0; i --)
     5         {
     6             if(digits[i] <= 8){
     7                 digits[i] += 1;
     8                 return digits;
     9             }else{//9
    10                 if(i != 0)
    11                     digits[i] = 0;
    12                 else
    13                 {
    14                     digits[0] = 1;
    15                     digits.push_back(0);
    16                     return digits;
    17                 }
    18             }
    19         }
    20     }
    21 };

    67- Add Binary

    Given two binary strings, return their sum (also a binary string).

    For example,
    a = "11"
    b = "1"
    Return "100".

    Solution:二进制加法,和为2进1,和为3进1留1;

    复制代码
     1 class Solution {
     2 public:
     3     string addBinary(string a, string b) {
     4         int sizea=a.size(),sizeb=b.size();
     5         if(sizea<sizeb)return addBinary(b,a);
     6         int n=sizea-sizeb;
     7         string helper(n,'0');
     8         b = helper + b;
     9         int carry=0;
    10         for(int i=sizea-1;i>=0;i--){
    11             int sum=(a[i]-'0')+(b[i]-'0')+carry;
    12             if(sum==0);
    13             else if(sum==1){
    14                 a[i]='1';
    15                 carry=0;
    16             }else if(sum==2){
    17                 a[i]='0';
    18                 carry=1;
    19             }else if(sum==3){
    20                 a[i]='1';
    21                 carry=1;
    22             }
    23         }
    24         if(carry==1)a='1'+a;
    25         return a;
    26     }
    27 };
    复制代码
  • 相关阅读:
    openwrt 更改 debug 等级(hostapd)
    openwrt 中procd
    openwrt增加串口登录需要密码
    openwrt设置语言的过程
    小程序感悟123
    如何用php实现分页效果
    如何利用h5将视频设置为背景
    关于js中定时器的返回值问题
    canvas二:绘制圆和其他曲线
    canvas一:基本认识
  • 原文地址:https://www.cnblogs.com/irun/p/4709507.html
Copyright © 2020-2023  润新知