标签：
Compare two version numbers version1 and version2.
If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.
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.
Here is an example of version numbers ordering:
0.1 < 1.1 < 1.2 < 13.37
分析：题意为 比较两个版本数字version1和version2，值得注意的是.并不是代表小数点，例如：2.5表示第二个一级第五个二级版本
思路：可以进行逐位比较，把字符串转化成整型。
注意考虑test case：0.1 与 0.01 结果是：0.1>0.01
代码如下：
class Solution {
public:
int compareVersion(string version1, string version2) {
int v1, v2;
int t1 = 0, t2 = 0;
while (t1 < version1.length() || t2 < version2.length()) {
v1 = 0;
while (t1 < version1.length()) {
if (version1[t1] == ‘.‘) {
++t1;
break;
}
v1 = v1 * 10 + (version1[t1] - ‘0‘);
++t1;
}
v2 = 0;
while (t2 < version2.length()) {
if (version2[t2] == ‘.‘) {
++t2;
break;
}
v2 = v2 * 10 + (version2[t2] - ‘0‘);
++t2;
}
if (v1 > v2) return 1;
if (v1 < v2) return -1;
}
return 0;
}
};
其他解法：
class Solution {
public:
int compareVersion(string version1, string version2) {
const char *p1 = version1.c_str()-1;
const char *p2 = version2.c_str()-1;
do{
int v1 = 0, v2 =0;
if (p1){
v1=atoi(p1+1);
p1 = strchr(p1+1, ‘.‘);
}
if (p2){
v2=atoi(p2+1);
p2 = strchr(p2+1, ‘.‘);
}
if (v1<v2) return -1;
if (v2<v1) return 1;
}while(p1||p2);
return 0;
}
};
或：
class Solution {
public:
int compareVersion(string version1, string version2) {
int n = version1.size(), m = version2.size();
for (int i = 0, j = 0; i < n || j < m; i++, j++) {
size_t p = 0;
int a = ((i >= n) ? 0 : stoi(version1.substr(i), &p));
i += p;
int b = ((j >= m) ? 0 : stoi(version2.substr(j), &p));
j += p;
if (a > b) return 1;
if (a < b) return -1;
}
return 0;
}
};
或：
class Solution {
public:
int compareVersion(string version1, string version2) {
istringstream iss1(version1), iss2(version2);
string token1, token2;
while(!iss1.eof() || !iss2.eof()) {
getline(iss1, token1, ‘.‘); getline(iss2, token2, ‘.‘);
if(stoi(token1) > stoi(token2)) return 1;
if(stoi(token1) < stoi(token2)) return -1;
token1 = token2 = "0";
}
return 0;
}
};
leetcode：Compare Version Numbers
标签：
原文地址：http://www.cnblogs.com/carsonzhu/p/4703268.html
