[LeetCode] 190. Reverse Bits 翻轉二進制位

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?

Related problem: Reverse Integer

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

將輸入轉換成2進制字符串，再翻轉並擴充到32位，再將此32位的二進制轉為無符號整數

int 在內存中以二進制形式存儲，占據32位。用一個 int 型整數 ans 來記錄結果，采用移位操作，因為：1. 注意到移位操作比乘2、除2操作效率更高，2. 移位操作很好地繞開了整型運算的溢出以及符號問題。在每次循環中：ans 每次左移一位，當 n 的最低位為1時，ans 的最低位就變為1，n 每次右移一位。總共進行32次循環。

Java: 每次只需移動1位

class Solution {
    public int reverseBits(int n) {  
        int ans = 0;  
        for (int i = 0; i < 32; i++) {  
            ans <<= 1;  
            if ((n & 1) == 1)  
                ans++;  
            n >>= 1;  
        }  
        return ans;  
    }
}

Java: 第 i 次循環移動 i 位

class Solution {
    public int reverseBits(int n) {  
        int ans = 0;  
        for (int i = 0; i < 32; i++)  
            ans |= ((n >> i) & 1) << (31 - i);  
        return ans;  
    } 
}

Python:

class Solution:
    # @param n, an integer
    # @return an integer
    def reverseBits(self, n):
        string = bin(n)
        if '-' in string:
            string = string[:3] + string[3:].zfill(32)[::-1]
        else:
            string = string[:2] + string[2:].zfill(32)[::-1]
        return int(string, 2)

Python:

class Solution:
    # @param n, an integer
    # @return an integer
    def reverseBits(self, n):
        result = 0
        for i in xrange(32):
            result <<= 1
            result |= n & 1
            n >>= 1
        return result　　

Python:

def reverseBits(n) :
	res = 0	
	while (n > 0) :
		res = res << 1
		if (n & 1 == 1) :
			res = res ^ 1
		n = n >> 1
		
	return res

Python:

def reverseBits2(n):
    res = 0
    while n > 0:
        res <<= 1
        res |= n & 1
        n >>= 1
        
    return res 　　

Python: 將n的二進制表示從低位到高位的值依次取出，逆序排列得到翻轉后的值。

class Solution(object):
    def reverseBits(self, n):
        """
        :type n: int
        :rtype: int
        """
        res = 0
        for i in xrange(32):
            res <<= 1
            res |= ((n >> i) & 1)
        return res　

Python: 利用Python的bin()函數

class Solution(object):
    def reverseBits(self, n):
        """
        :type n: int
        :rtype: int
        """
        b = bin(n)[:1:-1]
        return int(b + '0'*(32-len(b)), 2)　　

C++:

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t res = 0;
        for (int i = 0; i < 32; ++i) {
            if (n & 1 == 1) {
                res = (res << 1) + 1;
            } else {
                res = res << 1;
            }
            n = n >> 1;
        }
        return res;
    }
};　　

 

LeetCode中有關位操作的題：

[LeetCode] 191. Number of 1 Bits 二進制數1的個數

Repeated DNA Sequences 求重復的DNA序列

Single Number 單獨的數字

Single Number II 單獨的數字之二 

Grey Code 格雷碼 

  

類似題目：

[LeetCode] 191. Number of 1 Bits 二進制數1的個數

[LeetCode] 7. Reverse Integer 翻轉整數

 

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