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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"

    Note:

    1. Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes.
    2. Version strings do not start or end with dots, and they will not be two consecutive dots.

    time = O(n), space = O(n)

    class Solution {
        public int compareVersion(String version1, String version2) {
            String[] v1 = version1.split("\.");
            String[] v2 = version2.split("\.");
            
            int n = Math.max(v1.length, v2.length);
            for(int i = 0; i < n; i++) {
                int x = i < v1.length ? Integer.parseInt(v1[i]) : 0;
                int y = i < v2.length ? Integer.parseInt(v2[i]) : 0;
                if(x > y) {
                    return 1;
                } else if(x < y) {
                    return -1;
                } else {
                    continue;
                }
            }
            return 0;
        }
    }
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  • 原文地址:https://www.cnblogs.com/fatttcat/p/13645513.html
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