GuilinDev
Lc0190
05 August 2008
190 Reverse Bits
Reverse bits of a given 32 bits unsigned integer.
Note:
- Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer’s internal binary representation is the same, whether it is signed or unsigned.
- In Java, the compiler represents the signed integers using 2’s complement notation. Therefore, in Example 2 above, the input represents the signed integer
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-3
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-1073741825
Example 1:
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Input: n = 00000010100101000001111010011100
Output: 964176192 (00111001011110000010100101000000)
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.
Example 2:
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Input: n = 11111111111111111111111111111101
Output: 3221225471 (10111111111111111111111111111111)
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.
Constraints:
- The input must be a binary string of length
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32
Follow up: If this function is called many times, how would you optimize it?
分析
方法1，逐位颠倒，写个循环，循环32次每次右移取出一位。
方法2：分治，若要翻转一个二进制串，可以将其均分成左右两部分，对每部分递归执行翻转操作，然后将左半部分拼在右半部分的后面，即完成了翻转。由于左右两部分的计算方式是相似的，利用位掩码和位移运算，自底向上地完成这一分治流程。
对于递归的最底层，需要交换所有奇偶位：
取出所有奇数位和偶数位； 将奇数位移到偶数位上，偶数位移到奇数位上。 类似地，对于倒数第二层，每两位分一组，按组号取出所有奇数组和偶数组，然后将奇数组移到偶数组上，偶数组移到奇数组上。以此类推。
需要注意的是，在某些语言（如 Java）中，没有无符号整数类型，因此对 n 的右移操作应使用逻辑右移。
代码
逐个颠倒
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public class Solution {
public int reverseBits(int n) {
//定义返回结果
int res = 0;
for(int i = 0; i <= 31; i++){
//1 & (n >> i表示n右移i位，并取出其最后一位
//将取出的最后一位左移31-i位，存入返回结果中
res += ((1 & (n >> i)) << (31 - i));
}
return res;
}
}
// ===========================================================
// 时间复杂度：O(logn)。空间复杂度：O(1)。
public class Solution {
public int reverseBits(int n) {
int rev = 0;
for (int i = 0; i < 32 && n != 0; ++i) {
rev |= (n & 1) << (31 - i);
n >>>= 1;
}
return rev;
}
}
分治
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// 时间复杂度：O(1)。空间复杂度：O(1)。
public class Solution {
private static final int M1 = 0x55555555; // 01010101010101010101010101010101
private static final int M2 = 0x33333333; // 00110011001100110011001100110011
private static final int M4 = 0x0f0f0f0f; // 00001111000011110000111100001111
private static final int M8 = 0x00ff00ff; // 00000000111111110000000011111111
public int reverseBits(int n) {
n = n >>> 1 & M1 | (n & M1) << 1;
n = n >>> 2 & M2 | (n & M2) << 2;
n = n >>> 4 & M4 | (n & M4) << 4;
n = n >>> 8 & M8 | (n & M8) << 8;
return n >>> 16 | n << 16;
}
}