Leetcode刷题

485.Max Consecutive Ones

Description
Given a binary array, find the maximum number of consecutive 1s in this array.
给定一个二进制数组，找出这个数组中最大的连续1的个数

public class Solution {    public int findMaxConsecutiveOnes(int[] nums) {       int maxNumber = 0,number = 0;       int len = nums.length;       //遇0则变0，遇1则加1       for(int i = 0;i < len; i++){           if(nums[i] == 0){                number = 0;           }           else{                number++;                if(maxNumber < number)                    maxNumber = number;           }           }       return maxNumber;    }}public class Solution {    public int findMaxConsecutiveOnes(int[] nums) {       int maxNumber = 0,number = 0;         int len = nums.length;         //遇1则加1，遇0则变0         for(int i = 0;i < len; i++){             if(nums[i] == 1){                 number ++;             }             else{                 if(number > maxNumber){                     maxNumber = number;                 }                 number = 0;             }         }         //注意num[len - 1]如果是1的话没有执行给maxNumber赋值的语句         if(number >maxNumber){             maxNumber = number;         }         return maxNumber;    }}

461. Hamming Distance

Description
The Hamming distance between two integers is the number of positions at which the corresponding bits are different.
给定两个整数，求其对应二进制位上是不同数字的个数。

public int hammingDistance(int x, int y) {        int distance = 0;        int k = x ^ y;        //调用Integer类中的bitcount方法求一个整数的二进制中1的个数        distance = Integer.bitCount(k);        return distance;        }public int hammingDistance(int x, int y) {        int distance = 0;        int k = x ^ y;        System.out.println(k);        while(k != 0){        //和1相与，结果为1则最低位为1，结果为0则最低位为0            if( (k & 1) == 1)                distance ++;            k = k >> 1;//右移        }        return distance;}

//逐位异或public int hammingDistance(int x, int y) {      int distance = 0;        while(x !=0 || y!=0){            if(((x & 1) ^ (y & 1)) == 1){                distance ++;                    }            x = x >> 1;            y = y >> 1;        }        return distance;    }

 //（借鉴）利用递归 public int hammingDistance(int x, int y) {        if((x ^ y) == 0) return 0;        return (x ^ y) % 2 + hammingDistance(x >> 1, y >> 1);    }

感想：1. 二进制位运算要充分利用二进制运算符，~，&，|，^（异或，相同为0，不同为1）
2. 巧妙利用递归简化代码