Tcdian / keep

今天不想做,所以才去做。
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165. Compare Version Numbers #322

Open Tcdian opened 3 years ago

Tcdian commented 3 years ago

165. Compare Version Numbers

比较两个版本号 version1 和 version2。 如果 version1 > version2 返回 1,如果 version1 < version2 返回 -1, 除此之外返回 0

你可以假设版本字符串非空,并且只包含数字和 . 字符。

 . 字符不代表小数点,而是用于分隔数字序列。

例如,2.5 不是“两个半”,也不是“差一半到三”,而是第二版中的第五个小版本。

你可以假设版本号的每一级的默认修订版号为 0。例如,版本号 3.4 的第一级(大版本)和第二级(小版本)修订号分别为 34。其第三级和第四级修订号均为 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

Tcdian commented 3 years ago

Solution

/**
 * @param {string} version1
 * @param {string} version2
 * @return {number}
 */
var compareVersion = function(version1, version2) {
    let versionArr1 = version1.split('.');
    let versionArr2 = version2.split('.');
    let maxLen = versionArr1.length > versionArr2.length ? versionArr1.length : versionArr2.length;
    for (let i = 0; i < maxLen; i++) {
        if ((Number(versionArr1[i] || '0')) > (Number(versionArr2[i] || '0'))) {
            return 1;
        } else if ((Number(versionArr1[i] || '0')) < (Number(versionArr2[i] || '0'))) {
            return -1;
        }
    }
    return 0;
};
function compareVersion(version1: string, version2: string): number {
    let versionArr1 = version1.split('.');
    let versionArr2 = version2.split('.');
    let maxLen = versionArr1.length > versionArr2.length ? versionArr1.length : versionArr2.length;
    for (let i = 0; i < maxLen; i++) {
        if ((Number(versionArr1[i] || '0')) > (Number(versionArr2[i] || '0'))) {
            return 1;
        } else if ((Number(versionArr1[i] || '0')) < (Number(versionArr2[i] || '0'))) {
            return -1;
        }
    }
    return 0;
};