Number Complement:数字填充位操作

Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.

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: 5Output: 2Explanation: 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: 1Output: 0Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0.

思路：

朴素思路：按位判断，累加求和

class Solution {   public static int findComplement(int num) {        if(num == 0) return 1;        int factor = 1;        int result = 0;        while(num>0){            if((num&1)==0){                result += factor;            }            factor = factor<<1;            num = num>>1;        }        return result;    }}

思路二：求出和呆求数字一样长的全一数字，二者异或

int findComplement(int num) {    int mask = num;    mask |= mask >> 1;    mask |= mask >> 2;    mask |= mask >> 4;    mask |= mask >> 8;    mask |= mask >> 16;    return num ^ mask;}

有答案说使用highestOneBit(int i)，实际上，其源码是：

 /**     * Returns an {@code int} value with at most a single one-bit, in the     * position of the highest-order ("leftmost") one-bit in the specified     * {@code int} value.  Returns zero if the specified value has no     * one-bits in its two's complement binary representation, that is, if it     * is equal to zero.     *     * @return an {@code int} value with a single one-bit, in the position     *     of the highest-order one-bit in the specified value, or zero if     *     the specified value is itself equal to zero.     * @since 1.5     */    public static int highestOneBit(int i) {        // HD, Figure 3-1        i |= (i >>  1);        i |= (i >>  2);        i |= (i >>  4);        i |= (i >>  8);        i |= (i >> 16);        return i - (i >>> 1);    }

实际上和思路二本质相同。