Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.
Note:
- The given integer is guaranteed to fit within the range of a 32-bit signed integer.
- 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.
這道題給了我們一個數,讓我們求補數。通過分析題目彙總的例子,我們知道需要做的就是每個位翻轉一下就行了,但是翻轉的起始位置上從最高位的1開始的,前面的0是不能被翻轉的,所以我們從高往低遍歷,如果遇到第一個1了後,我們的flag就賦值爲true,然後就可以進行翻轉了,翻轉的方法就是對應位異或一個1即可,參見代碼如下:
解法一:
class Solution { public: int findComplement(int num) { bool start = false; for (int i = 31; i >= 0; --i) { if (num & (1 << i)) start = true; if (start) num ^= (1 << i); } return num; } };
由於位操作裏面的取反符號~本身就可以翻轉位,但是如果直接對num取反的話就是每一位都翻轉了,而最高位1之前的0是不能翻轉的,所以我們只要用一個mask來標記最高位1前面的所有0的位置,然後對mask取反後,與上對num取反的結果即可,參見代碼如下:
解法二:
class Solution { public: int findComplement(int num) { int mask = INT_MAX; while (mask & num) mask <<= 1; return ~mask & ~num; } };
再來看一種迭代的寫法,一行搞定碉堡了,思路就是每次都右移一位,並根據最低位的值先進行翻轉,如果當前值小於等於1了,就不用再調用遞歸函數了,參見代碼如下:
解法三:
class Solution { public: int findComplement(int num) { return (1 - num % 2) + 2 * (num <= 1 ? 0 : findComplement(num / 2)); } };
參考資料:
https://discuss.leetcode.com/topic/74627/3-line-c
https://discuss.leetcode.com/topic/74968/simple-java-one-line-solution
https://discuss.leetcode.com/topic/74642/java-1-line-bit-manipulation-solution