算法14-----位运算操作（1）

1、应用技巧1：与&【消除x（二进制）最后一位1】

def checkPowerof2(x):
    return x>0 and x&(x-1)==0

应用三：判断n是否为4的幂：

def isFour(n):
    m=1
    while m<n:
        m=m<<2
    return m==n

2、应用技巧2：子集

// Non Recursion
class Solution {
    /**
     * @param S: A set of numbers.
     * @return: A list of lists. All valid subsets.
     */
    public List<ArrayList<Integer>> subsets(int[] nums) {
        List<ArrayList<Integer>> result = new ArrayList<List<Integer>>();
        int n = nums.length;
        Arrays.sort(nums);
        
        // 1 << n is 2^n
        // each subset equals to an binary integer between 0 .. 2^n - 1
        // 0 -> 000 -> []
        // 1 -> 001 -> [1]
        // 2 -> 010 -> [2]
        // ..
        // 7 -> 111 -> [1,2,3]
        for (int i = 0; i < (1 << n); i++) {
            List<Integer> subset = new ArrayList<Integer>();
            for (int j = 0; j < n; j++) {
                // check whether the jth digit in i's binary representation is 1
                if ((i & (1 << j)) != 0) {
                    subset.add(nums[j]);
                }
            }
            result.add(subset);
        }
        
        return result;
    }
}

3、应用技巧3：异或

 

java代码：

public class Solution {
    public int singleNumber(int[] nums) {
        int ones = 0, twos = 0;
        for(int i = 0; i < nums.length; i++){
            ones = (ones ^ nums[i]) & ~twos;
            twos = (twos ^ nums[i]) & ~ones;
        }
        return ones;
    }
}

public class Solution {
    public int[] singleNumber(int[] nums) {
        //用于记录，区分“两个”数组
        int diff = 0;
        for(int i = 0; i < nums.length; i ++) {
            diff ^= nums[i];
        }
        //取最后一位1
        //先介绍一下原码，反码和补码
        //原码，就是其二进制表示（注意，有一位符号位）
        //反码，正数的反码就是原码，负数的反码是符号位不变，其余位取反
        //补码，正数的补码就是原码，负数的补码是反码+1
        //在机器中都是采用补码形式存
        //diff & (-diff)就是取diff的最后一位1的位置
        diff &= -diff;
        
        int[] rets = {0, 0}; 
        for(int i = 0; i < nums.length; i ++) {
            //分属两个“不同”的数组
            if ((nums[i] & diff) == 0) {
                rets[0] ^= nums[i];
            }
            else {
                rets[1] ^= nums[i];
            }
        }
        return rets;
    }
}