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

 

 

2、应用技巧2：子集

// Non Recursionclass Solution {    /**     * @param S: A set of numbers.     * @return: A list of lists. All valid subsets.     */    public List
     
      
       > subsets(int[] nums) {        List
       
        
         > result = new ArrayList
         
          
           >(); 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
           
             subset = new ArrayList
            
             (); 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;    }}