zoukankan      html  css  js  c++  java
  • LeetCode 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.

     1 /*************************************************************************
     2     > File Name: LeetCode198.c
     3     > Author: Juntaran
     4     > Mail: JuntaranMail@gmail.com
     5     > Created Time: Wed 11 May 2016 15:24:30 PM CST
     6  ************************************************************************/
     7  
     8 /*************************************************************************
     9     
    10     House Robber
    11     
    12     You are a professional robber planning to rob houses along a street. 
    13     Each house has a certain amount of money stashed, 
    14     the only constraint stopping you from robbing each of them 
    15     is that adjacent houses have security system connected 
    16     and it will automatically contact the police 
    17     if two adjacent houses were broken into on the same night.
    18 
    19     Given a list of non-negative integers representing 
    20     the amount of money of each house, 
    21     determine the maximum amount of money 
    22     you can rob tonight without alerting the police.
    23 
    24  ************************************************************************/
    25 
    26 #include <stdio.h>
    27 
    28 /* 改变了原数组的值 */
    29 int rob( int* nums, int numsSize )
    30 {
    31     if( numsSize == 0 )
    32     {
    33         return 0;
    34     }
    35     if( numsSize == 1 )
    36     {
    37         return nums[0];
    38     }
    39     
    40     nums[1] = nums[0]>nums[1] ? nums[0] : nums[1];
    41     
    42     int i;
    43     for( i=2; i<=numsSize-1; i++ )
    44     {
    45         nums[i] = (nums[i-2]+nums[i])>nums[i-1] ? (nums[i-2]+nums[i]) : nums[i-1];
    46     }
    47     return nums[numsSize-1];
    48 }
    49 
    50 /* 两个变量共同维护 */
    51 int rob2( int* nums, int numsSize )
    52 {
    53     if( numsSize == 0 )
    54     {
    55         return 0;
    56     }
    57     if( numsSize == 1 )
    58     {
    59         return nums[0];
    60     }
    61     
    62     int prev1 = 0;
    63     int prev2 = 0;
    64     
    65     int i, temp;
    66     for( i=0; i<=numsSize-1; i++ )
    67     {
    68         temp = prev1;
    69         prev1 = (prev2+nums[i])>prev1 ? (prev2+nums[i]) : prev1;
    70         prev2 = temp;
    71     }
    72     return prev1;
    73 }
    74 
    75 
    76 
    77 
    78 int main()
    79 {
    80     int nums[] = { 2,1,1,4,0 };
    81     int numsSize = 5;
    82     
    83     int ret = rob2( nums, numsSize );
    84     printf("%d
    ", ret);
    85     return 0;
    86 }
  • 相关阅读:
    forms组件、cookie与session
    choices参数、MTV与MVC模型、Ajax、序列化组件、sweetalert搭建页面、自定义分页器
    Django 常用字段,数据库查询优化 only与defer
    django配置代码
    django 模板层、模型层
    Django 请求生命周期、路由层
    centos6.8安装httpd后无法访问
    初次认识dedecms和帝国cms内容管理系统
    遇到一个json解码失败的问题
    关于apache配置映射端口
  • 原文地址:https://www.cnblogs.com/Juntaran/p/5482790.html
Copyright © 2011-2022 走看看