LeetCode—Compare Version Numbers，leetcodecompare

LeetCode—Compare Version Numbers，leetcodecompare

Compare two version numbers version1 and version1.
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

剛拿到題目的時候主要是對題目理解了很久啊

其實意思很簡單，每一個字串中有一些數字和‘.’，‘.’的作用只是為了把數字分開，然後依次比較大小

在自己寫的很土的方法中，就是簡單的先把兩個string分解到對應的vector當中，這裡有幾個需要注意的地方

兩個字串中的數字長度不是相等的，所以如果前面都相等還要看後面

11.22.33.00.00000 = 11.22.33

11.22.33.01 > 11.22.33

還有就是字串最後面是沒有'.'符號的，這裡需要注意

class Solution {public:    int compareVersion(string version1, string version2) {        vector<int> ver1;vector<int> ver2;int val1 = 0,val2 = 0;int i = 0,j = 0;while (i < version1.length()){if(version1[i] != '.'){val1 = val1*10+(version1[i]-'0');}else{ver1.push_back(val1);val1 = 0;}i++;}while (j < version2.length()){if (version2[j] != '.'){val2 = val2*10+(version2[j]-'0');}else{ver2.push_back(val2);val2 = 0;}j++;}ver1.push_back(val1);ver2.push_back(val2);bool flag = ver1.size() <= ver2.size();int length = flag? ver1.size():ver2.size();for (i = 0; i < length; i++){if (ver1[i] > ver2[i]){return 1;}else if(ver1[i] < ver2[i]){ return -1;}}if (flag){for (i = length; i < ver2.size(); i++){if (ver2[i] != 0){return -1;}}return 0;}else{for (i = length; i < ver1.size(); i++){if (ver1[i] != 0){return 1;}}return 0;}    }};

下面是網上的一些別的演算法，思想上都大同小異，但是代碼因為在while迴圈內，而且只迴圈一遍著實是簡便了很多：

class Solution {public:    int compareVersion(string version1, string version2) {        int lev1=0,lev2=0;        int id1=0,id2=0;        while(id1!=version1.length()||id2!=version2.length()){            lev1=0;            while(id1<version1.length()){                if(version1[id1]=='.'){                    ++id1;                    break;                }                lev1=lev1*10+(version1[id1]-'0');                ++id1;            }                        lev2=0;            while(id2<version2.length()){                if(version2[id2]=='.'){                    ++id2;                    break;                }                lev2=lev2*10+(version2[id2]-'0');                ++id2;            }                        if(lev1>lev2){                return 1;            }else if(lev1<lev2){                return -1;            }        }        return 0;//equal    }};