劍指offer程式設計題(JAVA實現)——第31題:整數中1出現的次數
阿新 • • 發佈:2018-12-27
github | https://github.com/JasonZhangCauc/JZOffer |
---|
- 劍指offer程式設計題(JAVA實現)——第31題:整數中1出現的次數
- 求出113的整數中1出現的次數,並算出1001300的整數中1出現的次數?
- 為此他特別數了一下1~13中包含1的數字有1、10、11、12、13因此共出現6次,但是對於後面問題他就沒轍了。
- ACMer希望你們幫幫他,並把問題更加普遍化,可以很快的求出任意非負整數區間中1出現的次數
- (從1 到 n 中1出現的次數)。
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;
}
/*
* 此方法空間不足
* 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;
}*/
}