Reverse Bits (E)
題目
Reverse bits of a given 32 bits unsigned integer.
Example 1:
Input: 00000010100101000001111010011100
Output: 00111001011110000010100101000000
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.
Example 2:
Input: 11111111111111111111111111111101
Output: 10111111111111111111111111111111
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10101111110010110010011101101001.
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 signed integer type and should not affect your implementation, as the internal binary representation of the integer 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
-3and the output represents the signed integer-1073741825.
Follow up:
If this function is called many times, how would you optimize it?
題意
將32位無符號整數的二進制數字逆置，回傳構成的新整數，
思路
每次取一位逆序處理，
代碼實作
Java
public class Solution {
// you need treat n as an unsigned value
public int reverseBits(int n) {
int ans = 0;
for (int i = 0; i < 32; i++) {
ans <<= 1;
ans |= n & 1;
n >>= 1;
}
return ans;
}
}
JavaScript
API
/**
* @param {number} n - a positive integer
* @return {number} - a positive integer
*/
var reverseBits = function(n) {
return Number.parseInt(n.toString(2).padStart(32, '0').split('').reverse().slice(0, 32).join(''), 2)
};
位運算
/**
* @param {number} n - a positive integer
* @return {number} - a positive integer
*/
var reverseBits = function (n) {
let ans = 0
for (let i = 0; i < 32; i++) {
ans <<= 1
ans |= n & 1
n >>= 1
}
ans >>>= 0 // js位運算默認使用有符號整數，該步驟可以將其轉化為無符號整數
return ans
}
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