题意的换一种说法:
将输入的两个数字,分别用.分割,然后逐一比较大小。如果数字数量不一样,多的大。前提是多出来的有不为0的。
class Solution { public: void getNums(string version1, vector<int> &v1) { int s = 0; int pos = version1.find("."); while (-1 != pos) { v1.push_back(atoi(version1.substr(s, pos - s).c_str())); s = pos + 1;</p> <pre><code> pos = version1.find(&quot;.&quot;, pos + 1); } v1.push_back(atoi(version1.substr(s).c_str())); } int compareVersion(string version1, string version2) { vector&lt;int&gt; v1, v2; v1.clear(); v2.clear(); getNums(version1, v1); getNums(version2, v2); int min = v1.size() &gt; v2.size() ? v2.size() : v1.size(); for (int i = 0;i &lt; min;i++) { if (v1[i] &gt; v2[i]) { return 1; } if (v1[i] &lt; v2[i]) { return -1; } } if (v1.size() &gt; v2.size()) { for (int i = min; i &lt; v1.size(); i++) { if (v1[i] &gt; 0) { return 1; } } return 0; } if (v1.size() &lt; v2.size()) { for (int i = min; i &lt; v2.size(); i++) { if (v2[i] &gt; 0) { return -1; } } return 0; } return 0; } </code></pre> <p>};