[LeetCode]--190. Reverse Bits（不是很懂的位运算）

补充知识，Java的位运算（bitwise operators）

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.

public int reverseBits(int n) {
        int reversed = 0;
        for (int i = 0; i < 32; i++) {
            reversed = (reversed << 1) | (n & 1);
            n = (n >> 1);
        }
        return reversed;
    }

懂了按位运算这个程序很容易看懂，但是很难想到。

我的思路是这样的。先转到32位二进制的String，然后交换之后又转到int。我觉得思路是正确的，但是有一些问题，不能通过。

public int reverseBits(int n) {
        String str = intToBinary(n);
        char[] strs = str.toCharArray();
        char temp = ' ';
        int i = 0, j = strs.length - 1;
        while (i < j) {
            temp = strs[i];
            strs[i] = strs[j];
            strs[j] = temp;
            i++;
            j--;
        }
        str = String.valueOf(strs);
        return binaryToInt(str);
    }

    public String intToBinary(int n) {
        String str = "";
        while (n != 0) {
            str = n % 2 + str;
            n = n / 2;
        }
        int len = str.length();
        for (int i = 0; i < 32 - len; i++) {
            str = "0" + str;
        }
        return str;
    }

    public int binaryToInt(String str) {
        int len = str.length() - 1;
        double sum = 0;
        for (int i = 0; i <= len; i++) {
            sum = sum + charToInt(str.charAt(i)) * Math.pow(2, len - i);
        }
        System.out.println(sum + "---" + Integer.MAX_VALUE);
        System.out.println(2147483647 + 1);
        return 0;
    }

    public int charToInt(char c) {
        if (c == '0')
            return 0;
        else
            return 1;
    }

第一个问题在于int会超出范围的。