LeetCode 165 Compare Version Numbers

思路

字符串根据"."得到每段的数字，依次进行比较即可。
follow up：
考虑尽可能多的corner case，判断输入的version是否合法。

复杂度

时间复杂度O(n), 空间复杂度O(1)

代码

class Solution {
    public int compareVersion(String version1, String version2) {
        // Compare each character in 2 strings.
        // The larger string wins
        // Longer string wins: one of the 2 strings hits the end, and they are the same so far. 
        
        int index1 = 0, index2 = 0;
        while(index1 < version1.length() || index2 < version2.length()) {
            int v1 = 0, v2 = 0;
            while(index1 < version1.length() && version1.charAt(index1) != '.') {
                v1 *= 10;
                v1 += version1.charAt(index1) - '0';
                index1++;
            }
            
            while(index2 < version2.length() && version2.charAt(index2) != '.') {
                v2 *= 10;
                v2 += version2.charAt(index2) - '0';
                index2++;
            }
            
            if(v1 < v2) return -1;
            if(v1 > v2) return 1;
            // or infinite loop occurs
            index1++;
            index2++;
        }
        return 0;
    }
}