題目

https://leetcode.com/problems/number-complement/

題解

class Solution {public int findComplement(int num) {int i = Integer.numberOfLeadingZeros(num);return ~num << i >> i;} }

也可以參考評(píng)論區(qū)的答案：

Idea:

• Flip all bits using negation operator (~) [read more here]
• Logical AND (&) with a bit a mask of size n whose all bits are set, where n = number of bits in num
  Example:

num = 5num binary = 0...0101~num binary = 1...1010* Now we've flipped all the bits but it also flipped previously insignificant bits (0s), so to revert them back to 0s, we need to only keep the significant bits and turn off the insignificant bits. * This can be done by using mask of 1s having size equal to number of bits in num. This mask is (1 << nBits) - 1. * This is a standard mask in bit manipulation problems..~num binary & mask = (1...1010) & (0...0111) = 0...0010 [Ans]

T/S: O(1)/O(1)

public int findComplement(int num) {var nBits = (int) Math.floor((Math.log(num) / Math.log(2)) + 1);var mask = (1 << nBits) - 1;return ~num & mask; }

1 liner variation of above method

public int findComplement(int num) {return ~num & (Integer.highestOneBit(num) - 1); }

總結(jié)

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