165. Compare Version Numbers

Given two version numbers, version1 and version2, compare them.

Version numbers consist of one or more revisions joined by a dot '.'. Each revision consists of digits and may contain leading zeros. Every revision contains at least one character. Revisions are 0-indexed from left to right, with the leftmost revision being revision 0, the next revision being revision 1, and so on. For example 2.5.33 and 0.1 are valid version numbers.

To compare version numbers, compare their revisions in left-to-right order. Revisions are compared using their integer value ignoring any leading zeros. This means that revisions 1 and 001 are considered equal. If a version number does not specify a revision at an index, then treat the revision as 0. For example, version 1.0 is less than version 1.1 because their revision 0s are the same, but their revision 1s are 0 and 1 respectively, and 0 < 1.

Return the following:

• If version1 < version2, return -1.
• If version1 > version2, return 1.
• Otherwise, return 0.

答案

0-indexed from left to right -> 从0排序，从左到右

func compareVersion(version1 string, version2 string) int {
	vs1 := strings.Split(version1, ".")
	vs2 := strings.Split(version2, ".")
	i, l1, l2 := 0, len(vs1), len(vs2)
	for ; i < l1 && i < l2; i++ {
		x, _ := strconv.Atoi(vs1[i])
		y, _ := strconv.Atoi(vs2[i])
		if x > y {
			return 1
		} else if x < y {
			return -1
		}
	}

	//走到这一步，意味着在前面的小版本比较中，都是相等的,而有一个版本已经走到了尽头
	if l1 == l2 {
		return 0
	}

	if l1 > l2 {
		if !isZero(vs1[i:]) {
			return 1
		}
	} else {
		if !isZero(vs2[i:]) {
			return -1
		}
	}

	return 0
}

func isZero(vs []string) bool {
	for i := range vs {
		num, _ := strconv.Atoi(vs[i])
		if num != 0 {
			return false
		}
	}
	return true
}