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[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.

 

这道题调试了好久,一直不想上网搜别人的解法,因为感觉自己可以做出来,改来改去最后终于通过了,再上网一搜,发现果然和别人的方法不同,小有成就感。我的思路是:由于两个版本号所含的小数点个数不同,有可能是1和1.1.1比较,还有可能开头有无效0,比如01和1就是相同版本,还有可能末尾无效0,比如1.0和1也是同一版本。对于没有小数点的数字,可以默认为最后一位是小数点,而版本号比较的核心思想是相同位置的数字比较,比如题目给的例子,1.2和13.37比较,我们都知道应该显示1和13比较,13比1大,所以后面的不用再比了,再比如1.1和1.2比较,前面都是1,则比较小数点后面的数字。那么算法就是每次对应取出相同位置的小数点之前所有的字符,把他们转为数字比较,若不同则可直接得到答案,若相同,再对应往下取。如果一个数字已经没有小数点了,则默认取出为0,和另一个比较,这样也解决了末尾无效0的情况。代码如下:

 

解法一:

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;
    }
};


当然我们也可以不使用将字符串转为整型的atoi函数,我们可以一位一位的累加,参加如下代码:

 

解法二:

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;
    }
};

 

由于这道题我们需要将版本号以’.'分开,那么我们可以借用强大的字符串流stringstream的功能来实现分段和转为整数,使用这种方法写的代码很简洁,如下所示:

 

解法三:

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;
    }
};

 

最后我们来看一种用C语言的字符串指针来实现的方法,这个方法的关键是用到将字符串转为长整型的strtol函数,关于此函数的用法可以参见我的另一篇博客strtol 函数用法。参见代码如下:

 

解法四:

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;
    }
};

 

类似题目:

First Bad Version

 

参考资料:

https://leetcode.com/problems/compare-version-numbers/discuss/?orderBy=most_votes

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

 

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2015-01-23 14:22  Grandyang  阅读(8853)  评论(3编辑  收藏
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