java中compareTo方法详解
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int java.lang.String.compareTo(String anotherString)Compares two strings lexicographically. The comparison is based on the Unicode value of each character in the strings. The character sequence represented by this String object is compared lexicographically to the character sequence represented by the argument string. The result is a negative integer if this String object lexicographically precedes the argument string. The result is a positive integer if this String object lexicographically follows the argument string. The result is zero if the strings are equal; compareTo returns 0 exactly when the equals(Object) method would return true.
This is the definition of lexicographic ordering. If two strings are different, then either they have different characters at some index that is a valid index for both strings, or their lengths are different, or both. If they have different characters at one or more index positions, let k be the smallest such index; then the string whose character at position k has the smaller value, as determined by using the < operator, lexicographically precedes the other string. In this case, compareTo returns the difference of the two character values at position k in the two string -- that is, the value:
this.charAt(k)-anotherString.charAt(k)
If there is no index position at which they differ, then the shorter string lexicographically precedes the longer string.
In this case, compareTo returns the difference of the lengths of the strings -- that is, the value:
this.length()-anotherString.length()
Specified by: compareTo(...) in Comparable
Parameters:
anotherString the String to be compared.
Returns:
the value 0 if the argument string is equal to this string; a value less than 0 if this string is lexicographically
less than the string argument; and a value greater than 0 if this string is lexicographically greater than the string
argument.
按字典序比较两个字符串,基于字符的Unicode编码比较。int c = s1.compareTo(s2);如果s1的字典序大于s2,比较结果值是负数;如果s1的字典序在s2的后面,比较结果值大于零;当且仅当s1与s2相等时,比较结果值为零。如果s1和s2在一个或多个位置处的字符不同,设k为最小的下标,则比较结果c值为:this.charAt(k)-anotherString.charAt(k)如果s1和s2没有字符不同的,则结果c值是两个字符串的长度的差值:this.length()-anotherString.length()
This is the definition of lexicographic ordering. If two strings are different, then either they have different characters at some index that is a valid index for both strings, or their lengths are different, or both. If they have different characters at one or more index positions, let k be the smallest such index; then the string whose character at position k has the smaller value, as determined by using the < operator, lexicographically precedes the other string. In this case, compareTo returns the difference of the two character values at position k in the two string -- that is, the value:
this.charAt(k)-anotherString.charAt(k)
If there is no index position at which they differ, then the shorter string lexicographically precedes the longer string.
In this case, compareTo returns the difference of the lengths of the strings -- that is, the value:
this.length()-anotherString.length()
Specified by: compareTo(...) in Comparable
Parameters:
anotherString the String to be compared.
Returns:
the value 0 if the argument string is equal to this string; a value less than 0 if this string is lexicographically
less than the string argument; and a value greater than 0 if this string is lexicographically greater than the string
argument.
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