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1丶丑数

因为丑数只能被2，3，5整除，所以让n依次去除2，3，5，如果最后是n为1，就说明是丑数。

class Solution {public boolean isUgly(int n) {if(n <= 0){return false;}while(n % 2 == 0){n /= 2;}while(n % 3 == 0){n /= 3;}while(n % 5 == 0){n /= 5;}return n == 1;}
}

2、各位相加

class Solution {public int addDigits(int num) {if(num<10)return num;while(num>=10){int sum=0;while (num > 0) {sum += num % 10;num /= 10;}num=sum;}return num;}
}

3丶搜索插入位置

本小菜坤的答案：

class Solution {public int searchInsert(int[] nums, int target) {for (int i = 0; i < nums.length; i++) {if (nums[i] == target) {return i;}}int place=0;for (int j = 0;  j< nums.length-1; j++) {if (nums[j] <= target && nums[j + 1] >= target) {return j+1;}}if(nums[nums.length-1]<target){    //判断最后一个元素return nums.length;}return 0;}}

4丶第一个错误的版本

看着题目挺复杂，其实思想就是二分查找的思想。

/* The isBadVersion API is defined in the parent class VersionControl.boolean isBadVersion(int version); */public class Solution extends VersionControl {public int firstBadVersion(int n) {int left=1;int right=n;while(left<right){int mid=left+(right-left)/2;  //中间值没有设置为(left+right)/2的原因是，当left和right都是非常大的整数时防止运算溢出if(isBadVersion(mid)){right=mid;}else{left=mid+1;}}return left;}
}