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[LeetCode] Complement of Base 10

[LeetCode] Complement of Base 10

作者: 埋没随百草 | 来源:发表于2019-04-21 16:07 被阅读0次

Every non-negative integer N has a binary representation. For example, 5 can be represented as "101" in binary, 11 as "1011" in binary, and so on. Note that except for N = 0, there are no leading zeroes in any binary representation.

The complement of a binary representation is the number in binary you get when changing every 1 to a 0 and 0 to a 1. For example, the complement of "101" in binary is "010" in binary.

For a given number N in base-10, return the complement of it's binary representation as a base-10 integer.

Example 1:

Input: 5
Output: 2
Explanation: 5 is "101" in binary, with complement "010" in binary, which is 2 in base-10.

Example 2:

Input: 7
Output: 0
Explanation: 7 is "111" in binary, with complement "000" in binary, which is 0 in base-10.

Example 3:

Input: 10
Output: 5
Explanation: 10 is "1010" in binary, with complement "0101" in binary, which is 5 in base-10.

Note:

0 <= N < 10^9

解题思路

第一种思路:遍历每一位,按位取反。(需记住0要特殊处理)
第二种思路:找到跟输入具有相同位数的最大整数max(把每一位都置为1),然后返回max & ~N即可。

实现代码

实现1:

// Runtime: 0 ms, faster than 100.00% of Java online submissions for Complement of Base 10 Integer.
// Memory Usage: 31.9 MB, less than 100.00% of Java online submissions for Complement of Base 10 Integer.
class Solution {
    public int bitwiseComplement(int N) {
        if (N == 0) {
            return 1;
        }

        int result = 0;
        int bit = 0;
        while (N > 0) {
            result |= (N % 2 == 0 ? 1 : 0) << bit++;
            N >>= 1;
        }
        return result;
    }
}

实现2:

// Runtime: 0 ms, faster than 100.00% of Java online submissions for Complement of Base 10 Integer.
// Memory Usage: 31.8 MB, less than 100.00% of Java online submissions for Complement of Base 10 Integer.
class Solution {
    public int bitwiseComplement(int N) {
        if (N == 0) {
            return 1;
        }

        int max = N;
        max |= max >>> 1;
        max |= max >>> 2;
        max |= max >>> 4;
        max |= max >>> 8;
        max |= max >>> 16;
        return max & ~N;
    }
}

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