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★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation. Note:
Example 1: Input: 5 Output: 2 Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2. Example 2: Input: 1 Output: 0 Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0. 给定一个正整数,输出它的补数。补数是对该数的二进制表示取反。 注意:
示例 1: 输入: 5 输出: 2 解释: 5的二进制表示为101(没有前导零位),其补数为010。所以你需要输出2。 示例 2: 输入: 1 输出: 0 解释: 1的二进制表示为1(没有前导零位),其补数为0。所以你需要输出0。 Runtime: 8 ms
Memory Usage: 3.8 MB
1 class Solution { 2 func findComplement(_ num: Int) -> Int { 3 return (1 - num % 2) + 2 * (num <= 1 ? 0 : findComplement(num / 2)) 4 } 5 } Runtime: 8 ms
Memory Usage: 3.8 MB
1 class Solution { 2 func findComplement(_ num: Int) -> Int { 3 var mask:Int = Int.max 4 while ((mask & num) != 0) 5 { 6 mask <<= 1 7 } 8 return ~mask & ~num 9 } 10 } 8ms 1 class Solution { 2 func findComplement(_ num: Int) -> Int { 3 var numberOfBitsNeeded = 0 4 var number = num 5 while number > 0 { 6 number /= 2 7 numberOfBitsNeeded += 1 8 } 9 10 11 let allOnes = (2 << (numberOfBitsNeeded - 1)) - 1 12 13 14 return (num ^ allOnes) 15 } 16 } 3842 kb1 class Solution { 2 func findComplement(_ num: Int) -> Int { 3 let numberOfBits = (String(num,radix:2)).count 4 return ((1 << numberOfBits) - 1) ^ num 5 } 6 } 3858 kb1 class Solution { 2 func findComplement(_ num: Int) -> Int { 3 4 let bitCount = num.bitWidth - num.leadingZeroBitCount 5 6 var mask = 1 7 8 for _ in 0..<bitCount - 1 { 9 mask = mask << 1 + 1 10 } 11 12 return num ^ mask 13 } 14 } 16ms 1 class Solution { 2 func findComplement(_ b: Int) -> Int { 3 var count = 0 4 var sum = 0 5 while sum <= b { 6 count += 1 7 sum = 1 << count 8 } 9 10 let rtn = sum - 1 - b 11 return rtn 12 } 13 } 28ms 1 class Solution { 2 func findComplement(_ num: Int) -> Int { 3 var num = num 4 var res = 0 5 var offset = 0 6 7 while num > 0 { 8 res = res + ((num & 1) ^ 1) << offset 9 num = num >> 1 10 offset += 1 11 } 12 13 return res 14 } 15 }
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