LeetCode - 273. Integer to English Words

最新推荐文章于 2024-07-30 07:35:52 发布

ryanylz

最新推荐文章于 2024-07-30 07:35:52 发布

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本文链接：https://blog.youkuaiyun.com/u014642924/article/details/68948182

本文介绍了一个将整数转换为其英文表达形式的C++程序。通过详细的代码解析，展示了如何处理不同位数及特殊数字（如11-19）的转换逻辑。

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解题代码:

classSolution {

public:

string find(int n){

switch(n){

case 1: return "One";

case 2: return "Two";

case 3: return "Three";

case 4: return "Four";

case 5: return "Five";

case 6: return "Six";

case 7: return "Seven";

case 8: return "Eight";

case 9: return "Nine";

default: return "";

}

string find2(int n){

switch(n){

case 2: return "Twenty";

case 3: return "Thirty";

case 4: return "Forty";

case 5: return "Fifty";

case 6: return "Sixty";

case 7: return "Seventy";

case 8: return "Eighty";

case 9: return "Ninety";

default: return "";

}

string find3(int n){

switch(n){

case 1: return "Eleven";

case 2: return "Twelve";

case 3: return"Thirteen";

case 4: return"Fourteen";

case 5: return "Fifteen";

case 6: return "Sixteen";

case 7: return"Seventeen";

case 8: return"Eighteen";

case 9: return"Nineteen";

default: return "";

}

string numberToWords(int num) {

string str;

if(num==0)

return "Zero";

if(num>=2000000000){

str+="Two Billion";

num-=2000000000;

if(num>0)

str+=" ";

}

if(num>=1000000000){

str+="One Billion";

num-=1000000000;

if(num>0)

str+=" ";

}

if(num>=1000000){

int a=num/1000000;

num%=1000000;

int c=a%10;

a/=10;

int b=a%10;

a/=10;

if(a!=0){

str+=find(a)+"Hundred";

if(b||c)

str+=" ";

}

if(b>=2){

str+=find2(b);

if(c!=0)

str+=" "+find(c);

}

if(b==1){

if(c!=0)

str+=find3(c);

else

str+="Ten";

}

if(b==0)

str+=find(c);

str+=" Million";

if(num>0)

str+=" ";

}

if(num>=1000){

int a=num/1000;

num%=1000;

int c=a%10;

a/=10;

int b=a%10;

a/=10;

if(a!=0){

str+=find(a)+"Hundred";

if(b||c)

str+=" ";

}

if(b>=2){

str+=find2(b);

if(c!=0)

str+=" "+find(c);

}

if(b==1){

if(c!=0)

str+=find3(c);

else

str+="Ten";

}

if(b==0)

str+=find(c);

str+=" Thousand";

if(num>0)

str+=" ";

}

if(num>=1){

int a=num;

int c=a%10;

a/=10;

int b=a%10;

a/=10;

if(a!=0){

str+=find(a)+"Hundred";

if(b||c)

str+=" ";

}

if(b>=2){

str+=find2(b);

if(c!=0)

str+=" "+find(c);

}

if(b==1){

if(c!=0)

str+=find3(c);

else

str+="Ten";

}

if(b==0)

str+=find(c);

}

return str;

}

};

解题思路:

题目要求返回数字的英文表达。骤眼看去并不难，只需按照数字从高位到低位开始逐个转换即可。但其中有不少容易出错的地方，如11~19的表达，以及空格的表达。容易在不适当的地方缺少或者多出空格，这就需要结合情况进行判断。