private final String[] HUNDRED = {"", "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"}; public String numberToWords(int num) { if(num == 0) return "Zero"; StringBuilder sb = new StringBuilder(); int index = 0; while(num > 0) { if(num...
999>0->result=hundred_to_word(999%1000=999/后三位传去函数)+result->num=999//1000=0(刮掉后三位)-> 0=0->‘’.join(result) 刮了3次,加后缀两次hundred_to_words+Thousands[2]+hundreds_to_words+Thousands[1]+hundred_to_words,i+=2 四、总结 从上可看出整个过程:1、只要num大于零就把后三...
主函数中调用四次这个帮助函数,然后中间要插入 "Thousand", "Million", "Billion" 到对应的位置,最后 check 一下末尾是否有空格,把空格都删掉,返回的时候检查下输入是否为0,是的话要返回 'Zero',参见代码如下: classSolution {public:stringnumberToWords(intnum) {stringres = convertHundred(num %1000); vecto...
Hint: Did you see a pattern in dividing the number into chunk of words? For example, 123 and 123000. Group the number by thousands (3 digits). You can write a helper function that takes a number less than 1000 and convert just that chunk to words. There are many edge cases. What a...
classSolution:defnumberToWords(self,num:int)->str:to19='One Two Three Four Five Six Seven Eight Nine Ten Eleven Twelve '\'Thirteen Fourteen Fifteen Sixteen Seventeen Eighteen Nineteen'.split()tens='Twenty Thirty Forty Fifty Sixty Seventy Eighty Ninety'.split()defhelper(num):ifnum<20:returnto...
class Solution{public:stringnumberToWords(intnum){if(!num)returnlow.back();string result="";for(inti=1000000000,j=0;j<4;i/=1000,j++){inttemp=num/i;if(temp){result+=helper(temp)+" "+high[j]+" ";num%=i;}}trim(result);returnresult;}private:vector<string>low{" ","One","Two"...
public String numberToWords(int num) { if(num<0) return ""; if(num==0) return "Zero"; String[] units = {""," Thousand"," Million"," Billion"}; String res=""; int i=0; while(num>0) { int one=num%1000; if(one>0) ...
Did you see a pattern in dividing the number into chunk of words? For example, 123 and 123000. Group the number by thousands (3 digits). You can write a helper function that takes a number less than 1000 and convert just that chunk to words. ...
atindex:Int,_numberToStr:[String])->String{guardindex>=0&&index<digits.countelse{fatalError("...
题目连接:https://leetcode.cn/problems/number-of-matching-subsequences/题目描述:给定字符串 s 和字符串数组 words , 返回words[i] 中是 s 的子序列的单词个数。 字符串的子序列是从原始字符串中生成的新字符串,可以从中删去一些字符(可以是none),而不改变其余字符的相对顺序。 例如, “ace” 是“abcde”...