LeetCode 273. 整数转英文单词的时间复杂度是多少？

如何解决LeetCode 273. 整数转英文单词的时间复杂度是多少？

我正在努力理解这个解决方案的时间复杂度。这个问题是关于将数字转换为英文单词。

例如， 输入：num = 1234567891 输出：“12345567891”

StringBuilder insert() 的时间复杂度为 O(n)。
我怀疑时间复杂度是 O(n^2)。但我不确定。其中 n 是位数。 问题链接：englishToWords

这是我的代码：

代码：

class Solution {
    private final String[] THOUSANDS = {"","Thousand","Million","Billion"};
    
    private final String[] LESS_THAN_TWENTY = {"","One","Two","Three","Four","Five","Six","Seven","Eight","Nine","Ten","Eleven","Twelve","Thirteen","Fourteen","Fifteen","Sixteen","Seventeen","Eighteen","Nineteen"};
    
    private final String[] TENS = {"","","Twenty","Thirty","Forty","Fifty","Sixty","Seventy","Eighty","Ninety"};
    //O(n) solution
    public String numberToWords(int num) {
        if (num == 0){
          return "Zero";  
        } 
        StringBuilder sb = new StringBuilder();
        int index = 0;
        //this contributes towards time complexity
        while (num > 0) {
            if (num % 1000 > 0) {
                StringBuilder tmp = new StringBuilder();
                helper(tmp,num % 1000);
                System.out.println(index);
                System.out.println("tmp: "+ tmp);
                tmp.append(THOUSANDS[index]).append(" ");
                //I suspect the time complexity will increase because of this to O(n^2)
                sb.insert(0,tmp);
            }
            index++;
            num = num / 1000;
        }
        return sb.toString().trim();
    }
    
    private void helper(StringBuilder tmp,int num) {
        if (num == 0) {
            return;
        } else if (num < 20) {
            tmp.append(LESS_THAN_TWENTY[num]).append(" ");
            return;
        } else if (num < 100) {
            tmp.append(TENS[num / 10]).append(" ");
            helper(tmp,num % 10);
        } else {
            tmp.append(LESS_THAN_TWENTY[num / 100]).append(" Hundred ");
            helper(tmp,num % 100);
        }
    }
}

解决方法

它是 O(log n)，假设 n 是 num，因为每个数字的数字少于 2 个单词n 的值，位数为 ceil(log₁₀(n))。

将值加倍可能会增加 2 个单词，仅此而已。

,

时间复杂度为 O(n)，其中 n 是给定数字中的位数。 我们正在遍历该数字一次。

我从在 Cisco 和 Google 工作的几个朋友那里证实了这一点。