[LeetCode 165] Compare Version Numbers

标签：java   leetcode   算法   

题目链接：compare-version-numbers

import java.util.ArrayList;
import java.util.List;

/**
 * 
		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
 *
 */

public class CompareVersionNumbers {

	
//	69 / 69 test cases passed.
//	Status: Accepted
//	Runtime: 237 ms
//	Submitted: 0 minutes ago
	
    static int compareVersion(String version1, String version2) {
    	
        List<Integer> v1 = new ArrayList<Integer>();
        List<Integer> v2 = new ArrayList<Integer>();        

        for (String s : version1.split("\\.")) {
			int n = 0;
			for (Character c : s.toCharArray()) 
				n = n * 10 + c - '0';
			v1.add(n);
		}
        for (String s : version2.split("\\.")) {
			int n = 0;
			for (Character c : s.toCharArray()) 
				n = n * 10 + c - '0';
			v2.add(n);
		}
        
        while(v1.size() != v2.size()) {
        	if(v1.size() < v2.size()) v1.add(0);
        	else v2.add(0);
        }
        
        for (int i = 0; i < Math.min(v1.size(), v2.size()); i++) {
			if(v1.get(i) < v2.get(i)) return -1;
			else if(v1.get(i) > v2.get(i)) return 1;
		}

        return 0;
        
    }
	public static void main(String[] args) {
		System.out.println(compareVersion("1.0", "1.10"));

	}

}

[LeetCode 165] Compare Version Numbers

标签：java   leetcode   算法   

原文地址：http://blog.csdn.net/ever223/article/details/44497891