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66 lines (61 loc) · 2.03 KB
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// 165. 比较版本号
/*
双指针：
1、同时遍历两个字符串，直到全部遍历完
2、双指针分别遍历两个字符串，获取逗号分隔的修订号，将字符转化成数字，比较两个修订号的大小，大于或小于则返回结果1或-1，相同则继续判断下一个修订号，
遍历结束后仍没有返回，说明修订号全部相同，返回0
12.01.33 12.11.33
↑ ↑ ==> 12 = 12
i j
===================================
12.01.33 12.11.33
↑ ↑ ==> 1 < 11
i j
*/
class Solution {
public int compareVersion(String version1, String version2) {
int n1 = version1.length(), n2 = version2.length();
int i = 0, j = 0;
while (i < n1 || j < n2) {
int num1 = 0, num2 = 0;
while (i < n1 && version1.charAt(i) != '.') {
num1 = num1 * 10 + version1.charAt(i++) - '0';
}
while (j < n2 && version2.charAt(j) != '.') {
num2 = num2 * 10 + version2.charAt(j++) - '0';
}
if (num1 > num2) {
return 1;
} else if (num1 < num2) {
return -1;
}
i++;
j++;
}
return 0;
}
}
/*
字符串分割：两个字符串根据'.'分割成字符串数组，同时遍历两个数组，把修订号转化成数字，分别比较修订号
*/
class Solution {
public int compareVersion(String version1, String version2) {
String[] s1 = version1.split("\\.");
String[] s2 = version2.split("\\.");
int n1 = s1.length, n2 = s2.length;
int i = 0, j = 0;
while (i < n1 || j < n2) {
int num1 = 0, num2 = 0;
if (i < n1) {
num1 = Integer.parseInt(s1[i++]);
}
if (j < n2) {
num2 = Integer.parseInt(s2[j++]);
}
if (num1 != num2) {
return num1 > num2 ? 1 : -1;
}
}
return 0;
}
}