求正则表达式的一个问题。
调用public String[] split(String regex,int limit)方法

例如，字符串 "boo:and:foo" 使用这些参数可生成下列结果：
  Regex Limit 结果
  o 5 { "b", "", ":and:f", "", "" }  

为什么b后面是一个空格,而":and:f"后面是两个空格?

------解决方案--------------------
String[] java.lang.String.split(String regex, int limit)
Splits this string around matches of the given regular expression. 

The array returned by this method contains each substring of this string that is terminated by another substring that matches the given expression or is terminated by the end of the string. The substrings in the array are in the order in which they occur in this string. If the expression does not match any part of the input then the resulting array has just one element, namely this string. 

The limit parameter controls the number of times the pattern is applied and therefore affects the length of the resulting array. If the limit n is greater than zero then the pattern will be applied at most n - 1 times, the array's length will be no greater than n, and the array's last entry will contain all input beyond the last matched delimiter. If n is non-positive then the pattern will be applied as many times as possible and the array can have any length. If n is zero then the pattern will be applied as many times as possible, the array can have any length, and trailing empty strings will be discarded. 

The string "boo:and:foo", for example, yields the following results with these parameters: 

Regex Limit Result 
: 2 { "boo", "and:foo" } 
: 5 { "boo", "and", "foo" } 
: -2 { "boo", "and", "foo" } 
o 5 { "b", "", ":and:f", "", "" } 
o -2 { "b", "", ":and:f", "", "" } 
o 0 { "b", "", ":and:f" } 

An invocation of this method of the form str.split(regex, n) yields the same result as the expression 

java.util.regex.Pattern. compile(regex).java.util.regex.Pattern.split(java.lang.CharSequence,int) split(str, n) 
See Also:
java.util.regex.Pattern
Parameters:
regex the delimiting regular expression
limit the result threshold, as described above
Returns:
the array of strings computed by splitting this string around matches of the given regular expression
Throws:
PatternSyntaxException if the regular expression's syntax is invalid
Since:
1.4
@spec
JSR-51
------解决方案--------------------
If the limit n is greater than zero then the pattern will be applied at most n - 1 times, the array 's length will be no greater than n, and the array 's last entry will contain all input beyond the last matched delimiter
------解决方案--------------------
如果limit n 大于 0，则模式将被最多应用 n - 1 次，数组的长度将不会大于 n，
[b][u]而且数组的最后一项将包含所有超出最后匹配的定界符的输入。[/u][/b] -->问题就在于这一句。
------解决方案--------------------
楼上正解
------解决方案--------------------
学习了 经测试终于理解
public static void main(String[] args) {
String[] cList1="boo:and:foo".split("o", 5);
System.out.println("开始");
for(int i = 0 ; i <cList1.length ;i++){
System.out.println(cList1[i]);
}
System.out.println("结束");
String[] cList2="boo:and:food".split("o", 5);
System.out.println("开始");