问题:git
Compare two version numbers version1 and version2.
If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.code
You may assume that the version strings are non-empty and contain only digits and the .
character.
The .
character does not represent a decimal point and is used to separate number sequences.
For instance, 2.5
is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.three
Here is an example of version numbers ordering:ci
0.1 < 1.1 < 1.2 < 13.37
解决:字符串
① 比较版本号,版本号通常表示为 2.10,3.2.3这样的只含有数字和小数点的形式。比较两个版本号的大小。【注】1和1.0是相等的。2.1表明2号大版本的第1个小版本,2.10表明2号大版本的第10个小版本,因此2.10比2.1大。同时,2.1.0和2.1表明的是同一个版本。string
class Solution { //3ms
public int compareVersion(String version1, String version2) {
String[] tmp1 = version1.split("\\.");
String[] tmp2 = version2.split("\\.");
int i = 0;
while(i < tmp1.length || i < tmp2.length){
if (i < tmp1.length && i < tmp2.length){
if (Integer.parseInt(tmp1[i]) < Integer.parseInt(tmp2[i])){
return -1;
}else if (Integer.parseInt(tmp1[i]) > Integer.parseInt(tmp2[i])){
return 1;
}
}else if(i < tmp1.length){
if (Integer.parseInt(tmp1[i]) != 0){
return 1;
}
}else if (i < tmp2.length){
if (Integer.parseInt(tmp2[i]) != 0){
return -1;
}
}
i ++;
}
return 0;
}
}it
② 将字符串转换为数字。io
class Solution {//0ms
public int compareVersion(String version1, String version2) {
if(version1 == null && version2 == null) return 0;
else if (version1 == null && version2 != null) return -1;
else if (version1 != null && version2 == null) return 1;
int len1 = version1.length();
int len2 = version2.length();
int tmp1 = 0;//用于暂存版本中的值
int tmp2 = 0;
int i1 = 0;
int i2 = 0;
while(i1 < len1 || i2 < len2){
while(i1 < len1 && version1.charAt(i1) != '.'){
tmp1 = 10 * tmp1 + (version1.charAt(i1) - '0');
i1 ++;
}
while(i2 < len2 && version2.charAt(i2) != '.'){
tmp2 = 10 * tmp2 + (version2.charAt(i2) - '0');
i2 ++;
}
if (tmp1 > tmp2) return 1;
else if(tmp1 < tmp2) return -1;
else{
tmp1 = 0;
tmp2 = 0;
i1 ++;
i2 ++;
}
}
return 0;
}
}class