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  • 算法: Reverse Bits 反转位收藏

    Reverse Bits

    Reverse bits of a given 32 bits unsigned integer.

    For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).

    Follow up: If this function is called many times, how would you optimize it?

    移位法

    复杂度

    时间 O(1) 空间 O(1)

    思路

    最简单的做法,原数不断右移取出最低位,赋给新数的最低位后新数再不断左移。

    代码

    public class Solution {
        // you need treat n as an unsigned value
        public int reverseBits(int n) {
            int res = 0;
            for(int i = 0; i < 32; i++, n >>= 1){
                res = res << 1 | (n & 1);
            }
            return res;
        }
    }

    分段相或法

    复杂度

    时间 O(1) 空间 O(1)

    思路

    Java标准的Integer.reverse()源码。

    代码

    public class Solution {
        // you need treat n as an unsigned value
        public int reverseBits(int i) {
            i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555;
            i = (i & 0x33333333) << 2 | (i >>> 2) & 0x33333333;
            i = (i & 0x0f0f0f0f) << 4 | (i >>> 4) & 0x0f0f0f0f;
            i = (i << 24) | ((i & 0xff00) << 8) | ((i >>> 8) & 0xff00) | (i >>> 24);
            return i;
        }
    }

    后续 Follow Up

    Q:如果该方法被大量调用,或者用于处理超大数据(Bulk data)时有什么优化方法?
    A:这其实才是这道题的精髓,考察的大规模数据时算法最基本的优化方法。其实道理很简单,反复要用到的东西记下来就行了,所以我们用Map记录之前反转过的数字和结果。更好的优化方法是将其按照Byte分成4段存储,节省空间。参见这个帖子。

     
    // cache
    private final Map<Byte, Integer> cache = new HashMap<Byte, Integer>();
    public int reverseBits(int n) {
        byte[] bytes = new byte[4];
        for (int i = 0; i < 4; i++) // convert int into 4 bytes
            bytes[i] = (byte)((n >>> 8*i) & 0xFF);
        int result = 0;
        for (int i = 0; i < 4; i++) {
            result += reverseByte(bytes[i]); // reverse per byte
            if (i < 3)
                result <<= 8;
        }
        return result;
    }
    
    private int reverseByte(byte b) {
        Integer value = cache.get(b); // first look up from cache
        if (value != null)
            return value;
        value = 0;
        // reverse by bit
        for (int i = 0; i < 8; i++) {
            value += ((b >>> i) & 1);
            if (i < 7)
                value <<= 1;
        }
        cache.put(b, value);
        return value;
    }
    转自:https://segmentfault.com/a/1190000003483740
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  • 原文地址:https://www.cnblogs.com/lwy1103/p/9025115.html
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