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  • 165. 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.

    You may assume the default revision number for each level of a version number to be 0. For example, version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0.

    Example 1:

    Input: version1 = "0.1", version2 = "1.1"
    Output: -1

    Example 2:

    Input: version1 = "1.0.1", version2 = "1"
    Output: 1

    Example 3:

    Input: version1 = "7.5.2.4", version2 = "7.5.3"
    Output: -1

    Example 4:

    Input: version1 = "1.01", version2 = "1.001"
    Output: 0
    Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”

    Example 5:

    Input: version1 = "1.0", version2 = "1.0.0"
    Output: 0
    Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"
    class Solution {
        public int compareVersion(String version1, String version2) {
            int i = 0, j = 0;
            int l1 = version1.length(), l2 = version2.length();
            
            while(i < l1 || j < l2){
                int c1 = 0, c2 = 0;
                while(i < l1 && version1.charAt(i) != '.'){
                    c1 = c1 * 10 + version1.charAt(i) - '0';
                    i++;
                }
                i++;
                while(j < l2 && version2.charAt(j) != '.'){
                    c2 = c2 * 10 + version2.charAt(j) - '0';
                    j++;
                }
                j++;
                if(c1 > c2) return 1;
                if(c1 < c2) return -1;
            }
            return 0;
        }
    }

    以dot为区分,每次比较dot中间的数值,因为每次比较前都会重新置零,所以如果有leading zero就可以自动排除。每次计算完后比较大小,否则最后返回0.

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  • 原文地址:https://www.cnblogs.com/wentiliangkaihua/p/11756531.html
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