class Solution {
public int countDigitOne(int n) {
return f(n);
}
private int f(int n ) {
if (n <= 0)
return 0;
String s = String.valueOf(n);
int high = s.charAt(0) - '0';
int pow = (int) Math.pow(10, s.length()-1);
int last = n - high*pow;
if (high == 1) {
return f(pow-1) + last + 1 + f(last);
} else {
return pow + high*f(pow-1) + f(last);
}
}
}
面试题43. 1~n整数中1出现的次数
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转载自blog.csdn.net/qq_42350785/article/details/106590449
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