leetcode：Compare Version Numbers

標籤：

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：題意為 比較兩個版本數字version1和version2，值得注意的是.並不是代表小數點，例如：2.5表示第二個一級第五個二級版本

思路：可以進行逐位比較，把字串轉化成整型。

注意考慮test case：0.1 與 0.01  結果是：0.1>0.01

代碼如下：

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