# leetcode解題：(Easy) 190. Reverse Bits

題目：https://leetcode.com/problems/reverse-bits/

描述：將32位元的unsigned int中的位元順序顛倒後回傳結果的值

解題思路：[詳細解答來源](https://leetcode.com/problems/reverse-bits/solutions/1232842/java-c-0ms-o-1-time-complexity-in-place-detailed-explanation/)，用分治法(divide and conquer)每次將處理範圍中左半與右半邊的位元值互換，對一個有`2^n`位元的數字總共只需要換`n`次即可

程式碼：

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        //Java bit shift operartor: >> (signed right shift)
        //Java bit shift operartor: >>> (unsigned right shift)
        //Java bit shift operartor: << (unsigned/signed left shift)

        n = ((n & 0xffff0000) >>> 16) | ((n & 0x0000ffff) << 16);
        n = ((n & 0xff00ff00) >>> 8) | ((n & 0x00ff00ff) << 8);
        n = ((n & 0xf0f0f0f0) >>> 4) | ((n & 0x0f0f0f0f) << 4);
        n = ((n & 0xcccccccc) >>> 2) | ((n & 0x33333333) << 2);
        n = ((n & 0xaaaaaaaa) >>> 1) | ((n & 0x55555555) << 1);

        return n;
    }
}

時間複雜度：O(1)
空間複雜度：O(1)

###### tags: `leetcode` `easy` `bitwise operate`