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  • [LeetCode] Compare Version Numbers

    Compare two version numbers version1 and version1.
    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.

    Here is an example of version numbers ordering:

    0.1 < 1.1 < 1.2 < 13.37

    逐位比较就好了,把字符串转化成整型,这样就可以忽略前置0的影响了。

     1 class Solution {
     2 public:
     3     int compareVersion(string version1, string version2) {
     4         int val1, val2;
     5         int idx1 = 0, idx2 = 0;
     6         while (idx1 < version1.length() || idx2 < version2.length()) {
     7             val1 = 0; 
     8             while (idx1 < version1.length()) {
     9                 if (version1[idx1] == '.') {
    10                     ++idx1;
    11                     break;
    12                 }
    13                 val1 = val1 * 10 + (version1[idx1] - '0');
    14                 ++idx1;
    15             }
    16             val2 = 0; 
    17             while (idx2 < version2.length()) {
    18                 if (version2[idx2] == '.') {
    19                     ++idx2;
    20                     break;
    21                 }
    22                 val2 = val2 * 10 + (version2[idx2] - '0');
    23                 ++idx2;
    24             }
    25             if (val1 > val2) return 1;
    26             if (val1 < val2) return -1;
    27         }
    28         return 0;
    29     }
    30 };
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  • 原文地址:https://www.cnblogs.com/easonliu/p/4218557.html
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