leetcode 342. Power of Four 解题报告
2016-06-25 21:41
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原题链接
首先把特殊数字处理了,即非正数和1
将要处理的数字num & (num-1)。这样可以得出其总共有几个1(二进制)
若只有最高位为1,那再判断其二进制0的总个数是否是2的倍数。(4的二进制是两个0)。
得出答案。
解题思路
刚开始题目理解错了,不过仔细审题后轻松解决首先把特殊数字处理了,即非正数和1
将要处理的数字num & (num-1)。这样可以得出其总共有几个1(二进制)
若只有最高位为1,那再判断其二进制0的总个数是否是2的倍数。(4的二进制是两个0)。
得出答案。
解题代码
public class Solution { public boolean isPowerOfFour(int num) { if(num<=0){ return false; } if(num ==1 ) return true; int a = num & (num-1); if(a==0&&(((Integer.toBinaryString(num)).length()-1)&1)==0) return true; else return false; } }
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