计算机Leetcode, again

Leetcode - Power of Four

2016-10-15  本文已影响10人  Richardo92

My code:

public class Solution {
    public boolean isPowerOfFour(int num) {
        return num > 0 && (num & (num - 1)) == 0 && (num & 0X55555555) != 0;
    }
}

reference:
https://discuss.leetcode.com/topic/42860/java-1-line-cheating-for-the-purpose-of-not-using-loops

Good solution without good explanation,it's easy to find that power of 4 numbers have those 3 common features.First,greater than 0.Second,only have one '1' bit in their binary notation,so we use x&(x-1) to delete the lowest '1',and if then it becomes 0,it prove that there is only one '1' bit.Third,the only '1' bit should be locate at the odd location,for example,16.It's binary is 00010000.So we can use '0x55555555' to check if the '1' bit is in the right place.With this thought we can code it out easily!

Anyway, Good luck, Richardo! -- 10/14/2016

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