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；
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 };