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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 };
[LeetCode] Compare Version Numbers
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