日期:2014-05-20 浏览次数:21020 次
public class Test1 {
public static void main(String[] args) {
int n = letter2Number("abc");
System.out.println(n);
}
public static int letter2Number(String letters) {
if(!letters.matches("[a-zA-Z]+")) {
throw new IllegalArgumentException("Format ERROR!");
}
char[] chs = letters.toLowerCase().toCharArray();
int result = 0;
for(int i = chs.length - 1, p = 1; i >= 0; i--) {
result += getNum(chs[i]) * p;
p *= 26;
}
return result;
}
private static int getNum(char c) {
return c - 'a' + 1;
}
}
------解决方案--------------------
String str = "abcdefghijklmnopqrstuvwxyz";
public int getNum(String oneChar){//传一个字母
return str.indexOf(oneChar)+1;
}
public int q26(int a){
if(a==0){
return 1;
}
else if(a==1){
return 26;
}
else{
return 26*q26(a-1);
}
}
public int changeToNum(String moreChar){//传的是一个字符串
int all = 0;
int p = 0;
char[] newOne = moreChar.toCharArray();
//挨个字母换成十进制数,然后再运算
for(int i = 0 ;i <newOne.length ; i++){
String tag = Character.toString(newOne[i]);
int num = getNum(tag);
all = all + num*q26(newOne.length-i-1);
}
return all;
}
------解决方案--------------------
那么麻烦?
JDK里面就有了,一句话:
ava.lang.Integer.parseInt(String "abc", 26);
int java.lang.Integer.parseInt(String s, int radix) throws NumberFormatException
Parses the string argument as a signed integer in the radix specified by the second argument. The characters in the string must all be digits of the specified radix (as determined by whether java.lang.Character.digit(char, int) returns a nonnegative value), except that the first character may be an ASCII minus sign '-' ('\u002D') to indicate a negative value. The resulting integer value is returned.
An exception of type NumberFormatException is thrown if any of the following situations occurs:
The first argument is null or is a string of length zero.
The radix is either smaller than java.lang.Character.MIN_RADIX or larger than java.lang.Character.MAX_RADIX.
Any character of the string is not a digit of the specified radix, except that the first character may be a minus sign '-' ('\u002D') provided that the string is longer than length 1.
The value represented by the string is not a value of type int.
Examples:
parseInt("0", 10) returns 0
parseInt("473", 10) returns 473
parseInt("-0", 10) returns 0
parseInt("-FF", 16) returns -255
parseInt("1100110", 2) returns 102
parseInt("2147483647", 10) returns 2147483647
parseInt("-2147483648", 10) returns -2147483648
parseInt("2147483648", 10) throws a NumberFormatException
parseInt("99", 8) throws a NumberFormatException
parseInt("Kona", 10) throws a NumberFormatException
parseInt("Kona", 27) returns 411787