[leetcode] Compare Version Numbers
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From : https://leetcode.com/problems/compare-version-numbers/
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
class Solution {public: int compareVersion(string version1, string version2) { int v1, v2, i=0, j=0, len1 = version1.size(), len2 = version2.size(); while(i<len1 || j<len2) { v1 = v2 = 0; while(i<len1 && version1[i]!='.') v1 = v1*10 + version1[i++]-'0'; while(j<len2 && version2[j]!='.') v2 = v2*10 + version2[j++]-'0'; if(v1!=v2) return (v1>v2)-(v1<v2);i++;j++; } return 0; }};
public class Solution { public int compareVersion(String version1, String version2) { for (int i = 0, j = 0; i < version1.length() || j < version2.length(); ++i, ++j) {int v1 = 0;int v2 = 0;while (i < version1.length() && version1.charAt(i) != '.') {v1 = v1 * 10 + version1.charAt(i++) - '0';}while (j < version2.length() && version2.charAt(j) != '.') {v2 = v2 * 10 + version2.charAt(j++) - '0';}if (v1 != v2) {return v1 > v2 ? 1 : -1;}}return 0; }}
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