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2016.5.15——leetcode-HappyNumber,House Robber

时间:2016-05-15 16:41:51      阅读:292      评论:0      收藏:0      [点我收藏+]

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leetcode:HappyNumber,House Robbe

1.Happy Number

这个题中收获2点:

1.拿到题以后考虑特殊情况,代码中考虑1和4,或者说<6的情况,动手算下。(可能要在代码一步步测试中发现,目前还不知道怎么知道这些特殊情况)

2.数字的每一位时,n%10,n/10。

  题目:

  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.

  Example: 19 is a happy number

  • 12 + 92 = 82
  • 82 + 22 = 68
  • 62 + 82 = 100
  • 12 + 02 + 02 = 1

  思路:

  我的思路:(作为你个程序末末,希望大家不要嘲笑)

    将数字用pow(x,2)来计算,但是不知道怎么写,好吧,跟没有思路一样.

  Leetcode/discuss中的思路:

    (n%10)*(n%10),n/10.

    Tip:在leetcode上每道题都有个discuss,从里面可以看到大家的思路及讨论,个人觉得很不错

  代码:

    有两个我认为写的比较简洁且易懂的代码。

  代码1:

 1 bool isHappy(int n) {
 2             int num=0;
 3             while(n!=1&&n!=4)         //1为初始值,如何平方都为1。4则进入死循环         
 4             {
 5                 while(n)
 6                 {
 7                     num += (n%10) * (n%10);    //循环将每位平方
 8                     n/=10;
 9                 }
10                 n=num;
11                 num=0;
12             }
13             return 1==n;
14         }            

代码2:

 1 class Solution {
 2 public:
 3     bool isHappy(int n) {
 4         while(n>6)
 5     {
 6         int next = 0;
 7         while(n)
 8             {
 9                  next+=(n%10)*(n%10);
10                  n/=10;
11              }
12         n = next;
13     }
14     return n==1;
15     }
16 };        

代码1是将数字“1”和“4”单独拿出来看,代码2则是直接从n大于6看。其实两者是一样的,可以自己用算下,“2”和”4“是一样的,”3“,”5“,”6“最终也会进入”2“,”4“循环。

22 =4 

42=8

82=16

12+62=37

32+72=58

52+82=89

82+92=145

12+42+52=42

42+22=20

22=4

 

32=9

92=81

82+12=65

62+52=61

62+12=37(回到22中的循环)

 

52=25

22+52=29

22+92=85

82+52=89(回到22中的循环)

 

62=36

32+62=45

42+52=41

42+12=17

12+72=50

52=25(回到52中的循环)

2.House Robber

题目收获:

  1. 如何用迭代
  2. 关于C++代码编辑器闪退的:

      1).在上述代码位置处加入语句 cin.get();

      2).同上述代码添加位置,加入语句 system("pause");

  题目:

  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.

  思路:

  我的思路:

    给定数组,求数组中不相邻的数字的最大和。但是不知道代码怎么写。

  Leetcode/discuss中的思路:

    跟我的相同,求数组中不相邻的数字的最大和。

  代码:

  贴出3段代码,前两个是写的简洁易懂的,后面一个是有问题的代码。

  代码1:C++代码

class Solution {
public:
    int rob(vector<int>& nums) { 
        int n = nums.size(), pre = 0, cur = 0;    //cur为i-1时的总和,pre为i-2的总和
        for (int i = 0; i < n; i++) {
            int temp = max(pre + nums[i], cur);    //如果pre+nums[i](即i-2时的总值)大于cur(即i-1时的总值),此时的总和值为pre+nums[i],否则为cur.
            pre = cur;
            cur = temp;
        }
        return cur;
    }
};

  代码2:c代码,但是思路非常清楚

 1 int max(int a, int b) {
 2      return a > b ? a : b;
 3  }
 4 
 5 int rob(int* nums, int numsSize) {
 6     int* d;
 7 
 8     d = (int*) malloc (numsSize * sizeof(nums[0]));
 9 
10     d[0] = nums[0];
11     d[1] = max (nums[0], nums[1]);
12 
13 // d[i] means the most value that can be robbed before the ith store. For each store,
  // we have two choice: rob or not rob: (1)if robbing, d[i] = d[i-2] + nums[i], for stores robbed cannot be connected. (2)if not robbing, d[i] = d[i-1]
14 15 for (int i = 2; i < numsSize; i++) { 16 d[i] = max(d[i-2]+nums[i], d[i-1]); 17 } 18 19 return d[numsSize-1]; 20 }

  代码3:错误思路

#define max(a, b) ((a)>(b)?(a):(b))
int rob(int num[], int n) {
    int a = 0;
    int b = 0;

    for (int i=0; i<n; i++)
    {
        if (i%2==0)     //以奇偶来求最大,奇数和偶数必然不相邻,但却不一定是最大       
        {
            a = max(a+num[i], b);
        }
        else
        {
            b = max(a, b+num[i]);
        }
    }

    return max(a, b);
}    

  代码3的错误原因奇偶不一定就是最大,比如a = { 1, 4, 2, 2, 4, 5, 9, 5 }这数组的最大为17(只选9,4,4),奇数为16,偶数16。可知看奇偶思路是错的。

  带主函数的:不会写主函数,不过还是自己吭哧吭哧出来一个主函数的代码,还不知道测试的边界条件是什么。

#include "stdafx.h"    //控制台头文件
#include "iostream"    //cin,cout输入输出的头文件
#include "vector"        //vector的头文件
#include "algorithm"    //代码中的max()函数的头文件,min()也是这个
using namespace std;    //C++必须写,还不知道为什么

class Solution
{
public:
    int rob(vector<int>& nums)
    {
        int cur = 0;
        int pre = 0;
        int temp = 0;

        for (size_t i = 0; i < nums.size(); i++)   //在visul studio中必须写错size_t不然会报错,但是leetcode上的测试不用,直接写成int就可以 
        {
            temp = max(pre + nums[i], cur);
            pre = cur;
            cur = temp;
        }
        return cur;
    }
};


int _tmain(int argc, _TCHAR* argv[])
{
    vector<int> a = { 1, 4, 2, 2, 4, 5, 9, 5 };    //测试的,本来想写成不固定的数组,数组由输入来的,暂时还不会
    int m;
    Solution solution;    //貌似必须这么写,不能直接引用
    m = solution.rob(a);
    cout << "the money is " << m << endl;
    cin.get();    //防止编辑器闪退
    return 0;
}

  

 

2016.5.15——leetcode-HappyNumber,House Robber

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原文地址:http://www.cnblogs.com/zhuzhu2016/p/5495315.html

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