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| 1 | +/** |
| 2 | + * Time Complexity: O(32) - Constant time |
| 3 | + * Space Complexity: O(1) - No extra space |
| 4 | + */ |
| 5 | +public class Solution { |
| 6 | + // you need treat n as an unsigned value |
| 7 | + public int reverseBits(int n) { |
| 8 | + int result = 0; |
| 9 | + |
| 10 | + for (int i = 0; i < 32; i++) { |
| 11 | + result = (result << 1) | (n & 1); |
| 12 | + n >>>= 1; // Use unsigned right shift |
| 13 | + } |
| 14 | + |
| 15 | + return result; |
| 16 | + } |
| 17 | +} |
| 18 | + |
| 19 | +// Alternative approach using string conversion |
| 20 | +class SolutionStringConversion { |
| 21 | + public int reverseBits(int n) { |
| 22 | + String binary = Integer.toBinaryString(n); |
| 23 | + |
| 24 | + // Pad with leading zeros to make it 32 bits |
| 25 | + while (binary.length() < 32) { |
| 26 | + binary = "0" + binary; |
| 27 | + } |
| 28 | + |
| 29 | + // Reverse the string |
| 30 | + StringBuilder reversed = new StringBuilder(binary).reverse(); |
| 31 | + |
| 32 | + // Convert back to integer |
| 33 | + return Integer.parseUnsignedInt(reversed.toString(), 2); |
| 34 | + } |
| 35 | +} |
| 36 | + |
| 37 | +// Alternative approach using iterative |
| 38 | +class SolutionIterative { |
| 39 | + public int reverseBits(int n) { |
| 40 | + int result = 0; |
| 41 | + |
| 42 | + for (int i = 0; i < 32; i++) { |
| 43 | + result = (result << 1) | (n & 1); |
| 44 | + n >>>= 1; |
| 45 | + } |
| 46 | + |
| 47 | + return result; |
| 48 | + } |
| 49 | +} |
| 50 | + |
| 51 | +// Alternative approach using while loop |
| 52 | +class SolutionWhileLoop { |
| 53 | + public int reverseBits(int n) { |
| 54 | + int result = 0; |
| 55 | + int i = 0; |
| 56 | + |
| 57 | + while (i < 32) { |
| 58 | + result = (result << 1) | (n & 1); |
| 59 | + n >>>= 1; |
| 60 | + i++; |
| 61 | + } |
| 62 | + |
| 63 | + return result; |
| 64 | + } |
| 65 | +} |
| 66 | + |
| 67 | +// Alternative approach using enhanced for loop |
| 68 | +class SolutionEnhancedForLoop { |
| 69 | + public int reverseBits(int n) { |
| 70 | + int result = 0; |
| 71 | + |
| 72 | + for (int i = 0; i < 32; i++) { |
| 73 | + result = (result << 1) | (n & 1); |
| 74 | + n >>>= 1; |
| 75 | + } |
| 76 | + |
| 77 | + return result; |
| 78 | + } |
| 79 | +} |
| 80 | + |
| 81 | +// Alternative approach using recursive |
| 82 | +class SolutionRecursive { |
| 83 | + public int reverseBits(int n) { |
| 84 | + return reverseBitsHelper(n, 0, 0); |
| 85 | + } |
| 86 | + |
| 87 | + private int reverseBitsHelper(int n, int result, int i) { |
| 88 | + if (i >= 32) { |
| 89 | + return result; |
| 90 | + } |
| 91 | + |
| 92 | + return reverseBitsHelper(n >>> 1, (result << 1) | (n & 1), i + 1); |
| 93 | + } |
| 94 | +} |
| 95 | + |
| 96 | +// Alternative approach using lookup table |
| 97 | +class SolutionLookupTable { |
| 98 | + private static final int[] REVERSE_TABLE = { |
| 99 | + 0x00, 0x80, 0x40, 0xC0, 0x20, 0xA0, 0x60, 0xE0, |
| 100 | + 0x10, 0x90, 0x50, 0xD0, 0x30, 0xB0, 0x70, 0xF0, |
| 101 | + 0x08, 0x88, 0x48, 0xC8, 0x28, 0xA8, 0x68, 0xE8, |
| 102 | + 0x18, 0x98, 0x58, 0xD8, 0x38, 0xB8, 0x78, 0xF8, |
| 103 | + 0x04, 0x84, 0x44, 0xC4, 0x24, 0xA4, 0x64, 0xE4, |
| 104 | + 0x14, 0x94, 0x54, 0xD4, 0x34, 0xB4, 0x74, 0xF4, |
| 105 | + 0x0C, 0x8C, 0x4C, 0xCC, 0x2C, 0xAC, 0x6C, 0xEC, |
| 106 | + 0x1C, 0x9C, 0x5C, 0xDC, 0x3C, 0xBC, 0x7C, 0xFC, |
| 107 | + 0x02, 0x82, 0x42, 0xC2, 0x22, 0xA2, 0x62, 0xE2, |
| 108 | + 0x12, 0x92, 0x52, 0xD2, 0x32, 0xB2, 0x72, 0xF2, |
| 109 | + 0x0A, 0x8A, 0x4A, 0xCA, 0x2A, 0xAA, 0x6A, 0xEA, |
| 110 | + 0x1A, 0x9A, 0x5A, 0xDA, 0x3A, 0xBA, 0x7A, 0xFA, |
| 111 | + 0x06, 0x86, 0x46, 0xC6, 0x26, 0xA6, 0x66, 0xE6, |
| 112 | + 0x16, 0x96, 0x56, 0xD6, 0x36, 0xB6, 0x76, 0xF6, |
| 113 | + 0x0E, 0x8E, 0x4E, 0xCE, 0x2E, 0xAE, 0x6E, 0xEE, |
| 114 | + 0x1E, 0x9E, 0x5E, 0xDE, 0x3E, 0xBE, 0x7E, 0xFE, |
| 115 | + 0x01, 0x81, 0x41, 0xC1, 0x21, 0xA1, 0x61, 0xE1, |
| 116 | + 0x11, 0x91, 0x51, 0xD1, 0x31, 0xB1, 0x71, 0xF1, |
| 117 | + 0x09, 0x89, 0x49, 0xC9, 0x29, 0xA9, 0x69, 0xE9, |
| 118 | + 0x19, 0x99, 0x59, 0xD9, 0x39, 0xB9, 0x79, 0xF9, |
| 119 | + 0x05, 0x85, 0x45, 0xC5, 0x25, 0xA5, 0x65, 0xE5, |
| 120 | + 0x15, 0x95, 0x55, 0xD5, 0x35, 0xB5, 0x75, 0xF5, |
| 121 | + 0x0D, 0x8D, 0x4D, 0xCD, 0x2D, 0xAD, 0x6D, 0xED, |
| 122 | + 0x1D, 0x9D, 0x5D, 0xDD, 0x3D, 0xBD, 0x7D, 0xFD, |
| 123 | + 0x03, 0x83, 0x43, 0xC3, 0x23, 0xA3, 0x63, 0xE3, |
| 124 | + 0x13, 0x93, 0x53, 0xD3, 0x33, 0xB3, 0x73, 0xF3, |
| 125 | + 0x0B, 0x8B, 0x4B, 0xCB, 0x2B, 0xAB, 0x6B, 0xEB, |
| 126 | + 0x1B, 0x9B, 0x5B, 0xDB, 0x3B, 0xBB, 0x7B, 0xFB, |
| 127 | + 0x07, 0x87, 0x47, 0xC7, 0x27, 0xA7, 0x67, 0xE7, |
| 128 | + 0x17, 0x97, 0x57, 0xD7, 0x37, 0xB7, 0x77, 0xF7, |
| 129 | + 0x0F, 0x8F, 0x4F, 0xCF, 0x2F, 0xAF, 0x6F, 0xEF, |
| 130 | + 0x1F, 0x9F, 0x5F, 0xDF, 0x3F, 0xBF, 0x7F, 0xFF |
| 131 | + }; |
| 132 | + |
| 133 | + public int reverseBits(int n) { |
| 134 | + return (REVERSE_TABLE[n & 0xFF] << 24) | |
| 135 | + (REVERSE_TABLE[(n >>> 8) & 0xFF] << 16) | |
| 136 | + (REVERSE_TABLE[(n >>> 16) & 0xFF] << 8) | |
| 137 | + (REVERSE_TABLE[(n >>> 24) & 0xFF]); |
| 138 | + } |
| 139 | +} |
| 140 | + |
| 141 | +// More concise version |
| 142 | +class SolutionConcise { |
| 143 | + public int reverseBits(int n) { |
| 144 | + int result = 0; |
| 145 | + for (int i = 0; i < 32; i++) { |
| 146 | + result = (result << 1) | (n & 1); |
| 147 | + n >>>= 1; |
| 148 | + } |
| 149 | + return result; |
| 150 | + } |
| 151 | +} |
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