LeetCode整数与字符的转换问题

整数转罗马数字

给定一个整数&＃xff0c;将其转为罗马数字。输入确保在 1 到 3999 的范围内。
示例 :输入: 3&＃xff0c;输出: “III”。
字符 数值&＃xff1a;I 1&＃xff0c;V 5&＃xff0c;X 10&＃xff0c;L 50&＃xff0c;C 100&＃xff0c;D 500&＃xff0c;M 1000

class Solution {
public:string intToRoman(int num) {int values[]&＃61;{1000,900,500,400,100,90,50,40,10,9,5,4,1};string reps[]&＃61;{"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};string res;for(int i&＃61;0; i<13; i&＃43;&＃43;){while(num>&＃61;values[i]){num -&＃61; values[i];res &＃43;&＃61; reps[i];}}return res;}
};


罗马数字转整数

给定一个罗马数字&＃xff0c;将其转换成整数。输入确保在 1 到 3999 的范围内。
示例 : 输入: “III”,输出: 3

class Solution {
public:int romanToInt(string s) {int values[] &＃61; { 1000,900,500,400,100,90,50,40,10,9,5,4,1 };string reps[] &＃61; { "M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I" };int res&＃61;0;int index &＃61; 0;int i &＃61; 0;while( i <13&&index};


整数转换英文表示

将非负整数转换为其对应的英文表示。可以保证给定输入小于 2^31 - 1 。
示例&＃xff1a;输入: 12345, 输出: “Twelve Thousand Three Hundred Forty Five”。

class Solution {
public:vector keys &＃61; {1000000000,1000000,1000,100,90,80,70,60,50,40,30,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1,0};vector values &＃61; {"Billion","Million","Thousand","Hundred","Ninety","Eighty","Seventy","Sixty","Fifty","Forty","Thirty","Twenty","Nineteen","Eighteen","Seventeen","Sixteen","Fifteen","Fourteen","Thirteen","Twelve","Eleven","Ten","Nine","Eight","Seven","Six","Five","Four","Three","Two","One","Zero"};string numberToWords(int num) {for (int i &＃61; 0; i <32; &＃43;&＃43;i) {int key &＃61; keys[i];if (num >&＃61; key) {if (num >&＃61; 100) {return numberToWords(num / key) &＃43; " " &＃43; values[i] &＃43; (num % key > 0 ? " " &＃43; numberToWords(num % key) : "");} else if (num >&＃61; 20) {return values[i] &＃43; (num % 10 > 0 ? " " &＃43; numberToWords(num % 10) : "");} else {return values[i];}}}return "";}
};


字符串相乘

给定两个以字符串形式表示的非负整数 num1 和 num2&＃xff0c;返回 num1 和 num2 的乘积&＃xff0c;它们的乘积也表示为字符串形式。

示例 1: 输入: num1 &＃61; “2”, num2 &＃61; “3”. 输出: “6”

class Solution {
public:string multiply(string num1, string num2) {if (num1 &＃61;&＃61; "0" || num2 &＃61;&＃61; "0") return "0";int n1 &＃61; num1.length() - 1;int n2 &＃61; num2.length() - 1;vector<int> mul(n1 &＃43; n2 &＃43; 2);
//编译器默认初始化为0 n位数和m位数相乘最多为m&＃43;n位数。
//乘数 num1 位数为 MM&＃xff0c;被乘数 num2 位数为 NN&＃xff0c; num1 x num2 结果 res 最大总位数为 M&＃43;N
//num1[i] x num2[j] 的结果为 tmp(位数为两位&＃xff0c;"0x","xy"的形式)&＃xff0c;其第一位位于 res[i&＃43;j]&＃xff0c;第二位位于 res[i&＃43;j&＃43;1]。
//11*12 &＃61;132 1和2相乘的结果为2的下标加1的下标&＃xff0c;假设1的下标代表2乘了几个10&＃xff08;0下标开始的话&＃xff09;for (int i &＃61; n1; i >&＃61; 0; i--)for (int j &＃61; n2; j >&＃61; 0; j--) { //11 12int bitmul &＃61; (num1[i] - &＃39;0&＃39;)*(num2[j] - &＃39;0&＃39;);//两个位上的数的乘积bitmul &＃43;&＃61; mul[i &＃43; j &＃43; 1];//先加低位&＃xff0c;判断是否有新的进位mul[i &＃43; j] &＃43;&＃61; bitmul / 10;//高位mul[i &＃43; j &＃43; 1] &＃61; bitmul % 10;//低位}//去掉前导0int i &＃61; 0;while (i < n1 &＃43; n2 &＃43; 1 && mul[i] &＃61;&＃61; 0)i&＃43;&＃43;;int index &＃61; i;string multi&＃61;string(n1&＃43;n2&＃43;2-i,&＃39;0&＃39;);for (; i < n1 &＃43; n2 &＃43; 2; i&＃43;&＃43;)multi[i-index]&＃61;(mul[i]&＃43;&＃39;0&＃39;);//string类型的append函数return multi;}
};


把字符串转换成整数

思路

1.功能测试

正数/复数/0


2.边界值测试

最大的正整数/最小的负整数&＃xff08;数据上下溢出&＃xff09;


class Solution {
public:/*全局变量*/enum {kValid,kInvalid}; // 枚举元素&＃xff08;kValid&＃61;0&＃xff0c;kInvalid&＃61;1&＃xff09;int g_nStatus &＃61; kValid; // 标记是否是非法输入/*功能函数*/int StrToInt(string str){g_nStatus &＃61; kInvalid; // 初始标记为非法输入long long num &＃61; 0; // 存储结果const char* cstr &＃61; str.c_str();// 指向字符数组的指针// 判断是否是空指针和空字符串""if( (cstr !&＃61; NULL) && (*cstr !&＃61; &＃39;\0&＃39;) ){// 处理符号位int minus &＃61; 1;if(*cstr &＃61;&＃61; &＃39;-&＃39;){minus &＃61; -1;cstr&＃43;&＃43;;}else if(*cstr &＃61;&＃61; &＃39;&＃43;&＃39;){minus &＃61; 1;cstr&＃43;&＃43;;}// 处理其余位while(*cstr !&＃61; &＃39;\0&＃39;){if(*cstr > &＃39;0&＃39; && *cstr < &＃39;9&＃39;){// string转int类型g_nStatus &＃61; kValid; // 标记为合法输入num &＃61; num*10 &＃43; (*cstr -&＃39;0&＃39;); // string转换为int类型cstr&＃43;&＃43;;// 数据上下溢出if( ((minus>0) && (num > 0x7FFFFFFF)) ||((minus<0) && (num > 0x80000000)) ){g_nStatus &＃61; kInvalid; // 如果溢出&＃xff0c;则标记为非法输入num &＃61; 0;break;}}else{g_nStatus &＃61; kInvalid;num &＃61; 0;break;}}if(g_nStatus &＃61;&＃61; kValid)num &＃61; num * minus;}cout<<(int)num<<endl;return (int)num;}
};