问题描述
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。