版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/as1072966956/article/details/83050043
github | https://github.com/JasonZhangCauc/JZOffer |
---|
- 剑指offer编程题(JAVA实现)——第31题:整数中1出现的次数
- 求出113的整数中1出现的次数,并算出1001300的整数中1出现的次数?
- 为此他特别数了一下1~13中包含1的数字有1、10、11、12、13因此共出现6次,但是对于后面问题他就没辙了。
- ACMer希望你们帮帮他,并把问题更加普遍化,可以很快的求出任意非负整数区间中1出现的次数
- (从1 到 n 中1出现的次数)。
public class Test31 {
public static void main(String[] args) {
System.out.println(NumberOf1Between1AndN_Solution(5));
System.out.println(NumberOf1Between1AndN_Solution(13));
System.out.println(NumberOf1Between1AndN_Solution(0));
System.out.println(NumberOf1Between1AndN_Solution(-1));
System.out.println(NumberOf1Between1AndN_Solution(1000));
System.out.println(NumberOf1Between1AndN_Solution(1));
}
public static int NumberOf1Between1AndN_Solution(int n) {
int count = 0;
for (int i = 0; i <= n; i++) {
String tmp = String.valueOf(i);
char[] array = tmp.toCharArray();
for (int j = 0; j < array.length; j++) {
if (array[j] == '1') {
count++;
}
}
}
return count;
}
/*
* 此方法空间不足
* public static int NumberOf1Between1AndN_Solution(int n) {
if (n < 0) {
return 0;
}
int count = 0;
int m = n;
if (n == 1) {
count++;
return count;
}
if (n == 0) {
return 0;
}
while (n >= 1) {
int tmp = n % 10;
if (tmp == 1) {
count++;
}
n /= 10;
}
count = count + NumberOf1Between1AndN_Solution(m - 1);
return count;
}*/
}