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  • Leetcode题解(22)

    66. Plus One

    题目

    这题很简单,直接代码:

     1 class Solution {
     2 public:
     3     vector<int> plusOne(vector<int> &digits) {
     4         // IMPORTANT: Please reset any member data you declared, as
     5         // the same Solution instance will be reused for each test case.
     6         int a = 1;
     7         vector<int> ans;
     8         vector<int>::iterator it;
     9         for(int i = digits.size() - 1;i >= 0;i--)
    10         {
    11             it = ans.begin();
    12             int b = (a + digits[i]) % 10;
    13             a = (a + digits[i]) / 10;
    14             ans.insert(it, b);
    15         }
    16         if(a != 0)
    17         {
    18             it = ans.begin();
    19             ans.insert(it, a);
    20         }
    21         
    22         return ans;
    23     }
    24 };

    ------------------------------------------------------------------------------------分割线--------------------------------------------------------------

    67. Add Binary

    题目

    直接代码

     1 class Solution {
     2   public:
     3       string addBinary(string a, string b) {
     4           int lenA,lenB;
     5           lenA = a.length();
     6           lenB = b.length();
     7           string res="";
     8           
     9 
    10           if (0 == lenA)
    11           {
    12               return b;
    13           }
    14           if (0 == lenB)
    15           {
    16               return a;
    17           }
    18           int ia=lenA-1,ib=lenB-1;
    19           int count=0,temp;//进位
    20           char c;
    21           while (ia>=0&&ib>=0)
    22           {
    23               temp = a[ia]-'0'+b[ib]-'0'+count;
    24               count = temp/2;
    25               c = temp%2+'0';
    26               res = c+res;
    27               ia--;
    28               ib--;
    29           }
    30 
    31           while (ia>=0)
    32           {
    33               temp = a[ia]-'0'+count;
    34               count = temp/2;
    35               c = temp%2+'0';
    36               res = c+res;
    37               ia--;
    38           }
    39 
    40           while (ib>=0)
    41           {
    42               temp = b[ib]-'0'+count;
    43               count = temp/2;
    44               c = temp%2+'0';
    45               res = c+res;
    46               ib--;
    47           }
    48           if(count != 0)
    49           {
    50               c=count+'0';
    51               res = c+res;
    52           }
    53           return res;
    54 
    55       }
    56   };
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  • 原文地址:https://www.cnblogs.com/LCCRNblog/p/5174979.html
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