Note:
1,The given integer is guaranteed to fit within the range of a 32-bit signed integer.
2,You could assume no leading zero bit in the integer’s binary representation.
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.
簡單來說就是要他的補數拉
本來我想的非常簡單
就是直接
return ~num就好
但是後來才發現因為他是用int 32bit表示
所以假設你是題目的範例5 => 101的話 他的補數會變成 前面多出29個1再來才是010
最後就是-6...
因此邏輯上需要將前面的0通通遮起來
作法如下
class Solution { public: int findComplement(int num) { unsigned int mask = ~0; //根據num的bit數調整mask while(num&mask) mask <<= 1; return ~mask & ~num; } };
沒有留言 :
張貼留言