?LeetCode刷題實(shí)戰(zhàn)190：顛倒二進(jìn)制位
Reverse bits of a given 32 bits unsigned integer.
題意
示例
示例 1：
輸入: 00000010100101000001111010011100
輸出: 00111001011110000010100101000000
解釋: 輸入的二進(jìn)制串 00000010100101000001111010011100 表示無(wú)符號(hào)整數(shù) 43261596，
?????因此返回 964176192，其二進(jìn)制表示形式為 00111001011110000010100101000000。
示例 2：
輸入：11111111111111111111111111111101
輸出：10111111111111111111111111111111
解釋：輸入的二進(jìn)制串 11111111111111111111111111111101 表示無(wú)符號(hào)整數(shù) 4294967293，
? 因此返回 3221225471 其二進(jìn)制表示形式為 10111111111111111111111111111111 。
解題
/**
?????* Returns the value obtained by reversing the order of the bits in the
?????* two's complement binary representation of the specified {@code?int}
?????* value.
?????*
?????* @param?i the value to be reversed
?????* @return?the value obtained by reversing order of the bits in the
?????* specified {@code?int} value.
?????* @since?1.5
?????*/
????public?static?int?reverse(int?i)?{
????????// HD, Figure 7-1
????????i = (i & 0x55555555) << 1?| (i >>> 1) & 0x55555555;
????????i = (i & 0x33333333) << 2?| (i >>> 2) & 0x33333333;
????????i = (i & 0x0f0f0f0f) << 4?| (i >>> 4) & 0x0f0f0f0f;
????????return?reverseBytes(i);
????}
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++) {
????????????res = (res << 1) + (n & 1);
????????????n >>= 1;
????????}
????????return?res;
????}
}