將非負(fù)整數(shù)轉(zhuǎn)換為其對(duì)應(yīng)的英文表示。可以保證給定輸入小于 231 - 1 。

示例 1:

輸入: 123
輸出: “One Hundred Twenty Three”
示例 2:

輸入: 12345
輸出: “Twelve Thousand Three Hundred Forty Five”
示例 3:

輸入: 1234567
輸出: “One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven”
示例 4:

輸入: 1234567891
輸出: “One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One”
讀法就是仿效123,456,789每次三個(gè)數(shù)這樣的讀。我們一開(kāi)始將可能用到的所有的單詞都存儲(chǔ)起來(lái)。而且這道題目不需要and了。這樣就簡(jiǎn)單很多了。注意要處理“000”這樣的時(shí)候。
代碼如下：

class Solution {private:string units[11] = {""," Thousand"," Million"," Billion"};string num0To9[11] = {"Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"};string num10To19[11] = {"Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};string num10To90[11] = {"Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};public:string fcs(int num){if(num==0) return "";else if(num>=1&&num<=9) return num0To9[num];else if(num>=10&&num<=19) return num10To19[num-10];else if(num>=20&&num<=99){if(num%10==0) return num10To90[num/10-1];else return num10To90[num/10-1]+" "+num0To9[num%10];}else{//三位數(shù)的情況string ans="";if(num%100==0) return num0To9[num/100]+" Hundred";else return num0To9[num/100]+" Hundred "+fcs(num%100);}}string numberToWords(int num) {if(num==0) return "Zero";else if(num>=1&&num<=9) return num0To9[num];else if(num>=10&&num<=19) return num10To19[num-10];else{string ans="";int cnt=0;while(num){string ss=units[cnt];string temp=fcs(num%1000);if(temp=="") ss="";//num%1000==0的情況else temp=temp+ss;if(ans.length()!=0&&num%1000!=0) ans=temp+" "+ans;//對(duì)于空格的處理，一開(kāi)始的時(shí)候是不需要空格的，另外就是“000”的時(shí)候是不需要空格的。else ans=temp+ans;num=(num-num%1000)/1000;//處理前三位cnt++;}return ans;}} };//130264790

努力加油a啊，(^o)/~

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