16.11.4

189. Rotate Array

Rotate an array of n elements to the right by k steps.
For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].
分析：将数组元素循环右移k个单位

public class Solution {    public void rotate(int[] nums, int k) {        /*I don't understand the role of k%=nums.length. Doesn't it change the value of k? I tried without it and i get index out of bound in the reverse function.        This is because that sometimes, k is larger than the length of array. For example nums = [1, 2, 3, 4, 5], k = 12, this means we only need to rotate the last two numbers. k = k % nums.length = 2;        */        //获取实际需要右移的位数        k%=nums.length;        reverse(nums,0,nums.length-1);        reverse(nums,0,k-1);        reverse(nums,k,nums.length-1);    }    //逆序    public static void reverse(int[] nums,int start,int end){        int temp=0;        for(int i=start,j=end;i<=j;i++,j--){            temp=nums[j];            nums[j]=nums[i];            nums[i]=temp;        }    }}

172. Factorial Trailing Zeroes

Given an integer n, return the number of trailing zeroes in n!.
分析：返回n！后面的尾0个数

public class Solution {    public int trailingZeroes(int n) {        int result=0;        //将n除以5后的结果累加        while(n>0){            n/=5;            result+=n;        }        return result;    }}

190. Reverse Bits

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).
分析：将32位无符号整数的二进制表示逆序（位操作）

public class Solution {    // you need treat n as an unsigned value    public int reverseBits(int n) {        int result=0;        for(int i=0;i<32;i++){            //先取出当前的最低位(n>>i&0x1)，取出后右移舍弃一位            //将当前位左移到指定位置实现逆序            result|=(n>>i&0x1)<<(31-i);        }        return result;    }}

191. Number of 1 Bits

Write a function that takes an unsigned integer and returns the number of ’1’ bits it has (also known as the Hamming weight).
For example, the 32-bit integer ’11’ has binary representation 00000000000000000000000000001011, so the function should return 3.
分析：将一个无符号整数用二进制表示，并且返回其中的位数为1的个数（位操作）

public class Solution {    // you need to treat n as an unsigned value    public int hammingWeight(int n) {        int count=0;        int temp=0;        for(int i=0;i<32;i++){            //判断每一位是否为1            temp=(n>>i&0x1);            if(temp==1)count++;        }        return count;    }}