回复：leetcode刷题贴

第十八题：263. Ugly Number
Write a program to check whether a given number is an ugly number.
Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 6, 8 are ugly while 14 is not ugly since it includes another prime factor 7.
Note that 1 is typically treated as an ugly number.
判断一个整形数是否是丑陋的数。丑陋的数字是指只包含2,3，5因数的数字。
特殊的，1作为一个丑陋的数字。
函数格式:
bool isUgly(int num) {
}

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确实没有什么人看，刚刚看了一下，还有51个题，这个周末如果没有事情的话，就把50个题全部贴上去。完结此贴。
贴吧果然是小白的地方。
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第十九题：202. Happy Number
Write an algorithm to determine if a number is "happy".
A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.
判断一个数是不是happyNumber。这种数字的定义是，各位计算平方和，最终结果为1.
例子：
19 is a happy number
1平方 + 9平方 = 82
8平方+ 2平方 = 68
6平方 + 8平方 = 100
1平方 + 0平方 + 0平方 = 1
函数格式：
bool isHappy(int n) {
}

最近项目比较忙。昨天没有更。

第二十题：231. Power of Two
Given an integer, write a function to determine if it is a power of two.
判定一个整形数是否是2的幂。
函数格式：
bool isPowerOfTwo(int n) {
}

一行代码搞定的。这个题换成二进制想。

第二十一题：326. Power of Three
Given an integer, write a function to determine if it is a power of three.
判定一个数是否是三的幂。不要使用循环。
函数格式：
bool isPowerOfThree(int n) {
}

一种方法是循环，一种方法是递归，一种方法是求对数然后在算次方判断是否相等。
但是我都没有写，我写了一个很投机的方法。不过平均时间还是没有上去。

第二十二题:21. Merge Two Sorted Lists
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
合并两个有序连边，合并之后的链表也需要有序。
函数格式：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
}

估计是看不懂，所以回帖的人很少。
哎，怪不得上次去招人的时候，头说尽量找博士。


两张图。很久没写c语言了，觉得全是ifelse，好不爽。

第二十三题：198. House Robber
You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.
Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.
你是一个贼。你要偷一条街上的房子。这些房子唯一的限制就是不能偷两个相邻的房子。给定一个非负整数列表，寻求可以偷到的最大金额。
函数格式：
int rob(int* nums, int numsSize) {
}