[LeetCode] 1009. Complement of Base 10 Integer 十进制整数的补码

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:

1. 0 <= N < 10^9
2. This question is the same as 476: https://leetcode.com/problems/number-complement/

这道题让求十进制整数的补码，我们都知道补码是二进制中的一个概念，就是每一位上的数都跟原数是相反的，这里所谓的十进制的补码，其实就是先转为二进制，再求补码，再转回十进制数。既然是求补码，最直接的方法就是从低到高，一位一位的取出来，然后进行取反。对于一个十进制数取其二进制数的最低位的方法，就是‘与’上一个1，然后取反，取反就是‘异或’上一个1，接着就是左移对应的位数，可以用一个变量i来统计当前处理到第几位了，然后直接左移i位就行了。之后N要右移一位，i自增1即可，参见代码如下：

解法一：

class Solution {
public:
    int bitwiseComplement(int N) {
        if (N == 0) return 1;
		int res = 0, i = 0;
		while (N > 0) {
			res += ((N & 1) ^ 1) << i;
			N >>= 1;
			++i;
		}
		return res;
    }
};

我们也可以使用一个 for 循环，这样可以写的更简洁一些，思路都是完全一样的，参见代码如下：

解法二：

class Solution {
public:
    int bitwiseComplement(int N) {
        if (N == 0) return 1;
		int res = 0, i = 0;
		for (int i = 0; N > 0; ++i, N >>= 1) {
			res += ((N & 1) ^ 1) << i;
		}
		return res;
    }
};

下面的这种解法用到了一些补码的性质，那就是一个二进制数加上其补码，则每个位置上都会变成1，且位数和原数相同。这样的话我们就可以先求出这个各位上全是1的数，然后再做减法，就可以得到补码了，参见代码如下：

解法三：

class Solution {
public:
    int bitwiseComplement(int N) {
        int X = 1;
        while (N > X) X = X * 2 + 1;
        return X - N;
    }
};

由于‘异或’的特点，在求得了各位上都是1的数字后，直接‘异或’上原数，也可以得到补码，参见代码如下：

解法四：

class Solution {
public:
    int bitwiseComplement(int N) {
        int X = 1;
        while (N > X) X = X * 2 + 1;
        return X ^ N;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1009

参考资料：

https://leetcode.com/problems/complement-of-base-10-integer/

https://leetcode.com/problems/complement-of-base-10-integer/discuss/256740/JavaC%2B%2BPython-Find-111.....1111-greater-N

https://leetcode.com/problems/complement-of-base-10-integer/discuss/613118/4-approaches%3A-bitwise-operation-math-formular-one-naive-simulation-and-bonus

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