【LeetCode】476. Number Complement (java实现)

原题链接

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

原题

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.

题目要求

题目要求为:给定一个非负整数,求出其complement number。所谓complement number,指对整数二进制最高位为1的位之后的所有位置取反,如5的二进制表示为00……00101,起最高位为1的位置是3,因此只对3之后的所有位置取反,得到00*00010,最后得出complement number为2。

解法

解法一:先求出最高位为1的位数,然后计算出掩码,再将原数据取反后和掩码求与,即得出最后结果。

public int findComplement(int num) {
int valid = 0; // 最高位为1的位数
int tmp = num;
while(tmp > 0) {
tmp /= 2;
valid++;
} return ~num & ((1 << valid) - 1);
}

其中,(1 << valid) - 1是二进制操作中一种常用的求掩码的方式,如valid为3,那么1<<valid二进制为1000,再减去1就得到了0000111。

Top解法:Top解法与解法一完全一致,只是使用了Java库实现的求出最高位为1的库函数而已。当然,代码更加简洁。

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

测试用例:

public static void main(String[] args) {
Solution s = new Solution();
assert(s.findComplement(5) == 2);
assert(s.findComplement(1) == 0);
assert(s.findComplement(0) == 0);
}
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