LeetCode Top Interview Questions 190. Reverse Bits (Java版; Easy)

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LeetCode Top Interview Questions 190. Reverse Bits (Java版; Easy)

题目描述

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 10111111111111111111111111111111.
 

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 -3 and the output represents the signed integer -1073741825.

第一次做; 一步到位:直接把二进制的每一位安排到最终位置上; 时间复杂度O(N), 空间复杂度O(1)

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++){
            //每次都处理二进制的最低位, 这样方便进行&操作
            int cur = n&1;
            //直接把二进制的最低位安排到最终位置上, 不过怎么安排左移和右移???? 蠢蛋, 不存在右移!! 
            res = res + (cur<<(31-i));
            //update
            n = n>>1;
        }
        return res;
    }
}

第一次做; 大体流程: 十进制转二进制字符串; 二进制字符串逆序; 二进制字符串转十进制; 时间复杂度O(N),空间复杂度O(N); 可以一步到位, 直接把每一位安排到最终的位置上

/*
题目中没描述溢出的场景
*/
public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        //十进制转二进制并逆序
        StringBuilder sb = new StringBuilder();
        for(int i=0; i<32; i++){
            sb.append(n&1);
            //update;因为是进行固定32次的循环, 所以右移时带不带符号位都不影响
            n = n>>1;
        }
        //二进制转十进制
        int res = 0;
        for(int i=0; i<32; i++){
            //核心: 整型和字符型运算, 我忘记把字符型转成整型了!!!!!
            res = res*2 + sb.charAt(i) - '0';
        }
        return res;
    }
}