0165. Compare Version Numbers (M)

Compare Version Numbers (M)

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.

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.

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题意

比较两个版本号的大小。

思路

比较方便的是先以"."为分隔符将字符串拆成一个字符串数组，再挨个比较每一个字符串代表的数字大小。

当然也可以不用java内置的api实现，直接遍历字符串计算每一层级版本对应的数字。

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代码实现

Java

split

class Solution {
    public int compareVersion(String version1, String version2) {
        // 注意正则需要用"\\."来匹配"."
        String[] v1 = version1.split("\\.");
        String[] v2 = version2.split("\\.");
        int i = 0;

        while (i < v1.length || i < v2.length) {
            // 长度不够则当前对应数字为0
            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;
            if (x > y) return 1;

            i++;
        }
        
        return 0;
    }
}

遍历

class Solution {
    public int compareVersion(String version1, String version2) {
        int i = 0, j = 0;
        while (i < version1.length() || j < version2.length()) {
            int x = 0, y = 0;		// 长度不够时则为默认值0
            
            while (i < version1.length() && version1.charAt(i) != '.') {
                x = x * 10 + version1.charAt(i++) - '0';
            }
            i++;		// 移动到"."后一位数字
            while (j < version2.length() && version2.charAt(j) != '.') {
                y = y * 10 + version2.charAt(j++) - '0';
            }
            j++;		// 移动到"."后一位数字

            if (x < y) return -1;
            if (x > y) return 1;
        }
        return 0;
    }
}

JavaScript

/**
 * @param {string} version1
 * @param {string} version2
 * @return {number}
 */
var compareVersion = function (version1, version2) {
  let v1 = version1.split('.').map(v => parseInt(v))
  let v2 = version2.split('.').map(v => parseInt(v))
  let i = 0
  while (i < v1.length || i < v2.length) {
    let a = i < v1.length ? v1[i] : 0
    let b = i < v2.length ? v2[i] : 0
    if (a !== b) {
      return a < b ? -1 : 1
    }
    i++
  }
  return 0
}