leetcode [#165] | GCidea's blog

JerryXia 发表于 2018-04-13 14:14:14 , 阅读 (8)

1. 题目
2. 解决方案
3. 注意事项

题目

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
Here is an example of version numbers ordering:
0.1 < 1.1 < 1.2 < 13.37

解决方案

public class Solution {
    public int compareVersion(String version1, String version2) {
        String[] arr1 = version1.split("\\.");
        String[] arr2 = version2.split("\\.");
        List<String> list1 = new ArrayList<>();
        List<String> list2 = new ArrayList<>();
        for(int i = 0; i < arr1.length; i++){
            list1.add(arr1[i]);
        }
        for(int i = 0; i < arr2.length; i++){
            list2.add(arr2[i]);
        }
        int b1 = list1.size();
        int b2 = list2.size();
        if(b1 > b2){
            for(int k = 0; k < (b1 - b2); k++){
                list2.add("0");
            }
        }
        if(b1 < b2){
            for(int k = 0; k < (b2 - b1); k++){
                list1.add("0");
            }
        }
        int result = 0;
        for(int m = 0; m < list1.size(); m++){
            if(Integer.valueOf(list1.get(m)) > Integer.valueOf(list2.get(m))){
                result = 1;
                break;
            }
            if(Integer.valueOf(list1.get(m)) < Integer.valueOf(list2.get(m))){
                result = -1;
                break;
            }
        }
        return result;
    }
}

注意事项

先将字符串按“.”分割。
将分割后数组每个元素装入ArrayList并补齐长度。
依次比较两个list中的每个元素即可。

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