【leetcode】Ugly Number II Java
2016-12-22 15:41
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这道题在Ugly Number的基础上,求第N个Ugly Number
开始我的思路是:先写出算法去判断这个数是不是丑数,然后利用for循环,范围从1到Integer.MAX_VALUE,如
果这个数是丑数,则count++(count初始值为0), 当count == n,时,则找到了第n个丑数。我是这么想的,也是这么去
实施的,于是快快的写出代码,代码如下:
一运行,就出现下面的错误——超时,看来我还是太naive了。
正确解法如下所示:
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