# [LeetCode] Compare Version Numbers 版本比较

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.

Example 1:

Input: version1 = "0.1", version2 = "1.1"
Output: -1

Example 2:

Input: version1 = "1.0.1", version2 = "1"
Output: 1

Example 3:

Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

class Solution {
public:
int compareVersion(string version1, string version2) {
int n1 = version1.size(), n2 = version2.size();
int i = 0, j = 0, d1 = 0, d2 = 0;
string v1, v2;
while (i < n1 || j < n2) {
while (i < n1 && version1[i] != '.') {
v1.push_back(version1[i++]);
}
d1 = atoi(v1.c_str());
while (j < n2 && version2[j] != '.') {
v2.push_back(version2[j++]);
}
d2 = atoi(v2.c_str());
if (d1 > d2) return 1;
else if (d1 < d2) return -1;
v1.clear(); v2.clear();
++i; ++j;
}
return 0;
}
};

class Solution {
public:
int compareVersion(string version1, string version2) {
int n1 = version1.size(), n2 = version2.size();
int i = 0, j = 0, d1 = 0, d2 = 0;
while (i < n1 || j < n2) {
while (i < n1 && version1[i] != '.') {
d1 = d1 * 10 + version1[i++] - '0';
}
while (j < n2 && version2[j] != '.') {
d2 = d2 * 10 + version2[j++] - '0';
}
if (d1 > d2) return 1;
else if (d1 < d2) return -1;
d1 = d2 = 0;
++i; ++j;
}
return 0;
}
};

class Solution {
public:
int compareVersion(string version1, string version2) {
istringstream v1(version1 + "."), v2(version2 + ".");
int d1 = 0, d2 = 0;
char dot = '.';
while (v1.good() || v2.good()) {
if (v1.good()) v1 >> d1 >> dot;
if (v2.good()) v2 >> d2 >> dot;
if (d1 > d2) return 1;
else if (d1 < d2) return -1;
d1 = d2 = 0;
}
return 0;
}
};

class Solution {
public:
int compareVersion(string version1, string version2) {
int res = 0;
char *v1 = (char*)version1.c_str(), *v2 = (char*)version2.c_str();
while (res == 0 && (*v1 != '\0' || *v2 != '\0')) {
long d1 = *v1 == '\0' ? 0 : strtol(v1, &v1, 10);
long d2 = *v2 == '\0' ? 0 : strtol(v2, &v2, 10);
if (d1 > d2) return 1;
else if (d1 < d2) return -1;
else {
if (*v1 != '\0') ++v1;
if (*v2 != '\0') ++v2;
}
}
return res;
}
};

https://leetcode.com/problems/compare-version-numbers/discuss/50774/Accepted-small-Java-solution.

https://leetcode.com/problems/compare-version-numbers/discuss/50788/My-JAVA-solution-without-split

https://leetcode.com/problems/compare-version-numbers/discuss/50804/10-line-concise-solution.-(C%2B%2B)

https://leetcode.com/problems/compare-version-numbers/discuss/50767/My-2ms-easy-solution-with-CC%2B%2B

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posted @ 2015-01-23 14:22  Grandyang  阅读(9527)  评论(3编辑  收藏  举报