classSolution { public: string addStrings(string num1, string num2) { string res =""; intm = num1.size(), n = num2.size(), i = m - 1, j = n - 1, carry = 0; while(i >= 0 || j >= 0) { inta = i >= 0 ? num1[i--] -'0': 0; intb = j >= 0 ? num2...
classSolution{public:// 大整数加法stringaddStrings(stringnum1,stringnum2){stringn1(num1.rbegin(), num1.rend());stringn2(num2.rbegin(), num2.rend());intsize1 = n1.length();intsize2 = n2.length();if(size1 < size2) {for(inti =0; i < size2 - size1; i++) { n1 +='0';...
Add Strings: Given two non-negative integers, num1 and num2 represented as string, return the sum of num1 and num2 as a string. Example: Input: num1 = "456", num2 = "77" Output: "533" Solution The basic idea is to add the numbers one by one from tail to head. Code class ...
class Solution { public: string addStrings(string num1, string num2) { int len1 = num1.length(); int len2 = num2.length(); int temp = 0, flag = 0; int len = max(len1, len2); string s(len,'\0'); for (int i = 0; i<min(len1, len2); i++){ //相同位计算 temp ...
leetcode 415. Add Strings Given two non-negative integersnum1andnum2represented as string, return the sum ofnum1andnum2. Note: The length of bothnum1andnum2is < 5100. Bothnum1andnum2contains only digits0-9. Bothnum1andnum2does not contain any leading zero....
usestd::iter; implSolution{ pubfnadd_binary(a:String,b:String)->String{ //进位,初始为0 letmutcarry=0; //收集a和b按位加的结果 letmutresult=a //返回底层的字节切片.as_bytes() //转换成迭代器 .iter() //提前反向迭代(后面会加上无限的'0',所以不能后面同时反向) .rev() //在a后面串...
classSolution {public:stringaddBinary(stringa,stringb) {intm = a.size(),n =b.size();if(m <=0)returnb;if(n <=0)returna; m--,n--;stringres ="";intcarry =0;while(m >=0|| n >=0){intnum1 = m >=0? (a[m--] -'0') :0;intnum2 = n >=0? (b[n--] -'0') ...
class Solution { public ArrayList<String> answer; public String digits; public long target; public void recurse( int index, long previousOperand, long currentOperand, long value, ArrayList<String> ops) { String nums = this.digits; // Done processing all the digits in num ...
1449. 数位成本和为目标值的最大数字 - 给你一个整数数组 cost 和一个整数 target 。请你返回满足如下规则可以得到的 最大 整数: * 给当前结果添加一个数位(i + 1)的成本为 cost[i] (cost 数组下标从 0 开始)。 * 总成本必须恰好等于 target 。 * 添加的数位中没有
Leetcode-Add Binary Given two binary strings, return their sum (also a binary string). For example, a = "11" b = "1" Return "100". 二进制相加,字符串输入,字符串输出 public class Solution { public String addBinary(String a, String b) { ...