LeetCode 213

House Robber II

Note: This is an extension of House Robber.

After robbing those houses on that street,
the thief has found himself a new place for his thievery
so that he will not get too much attention.
This time, all houses at this place are arranged in a circle.
That means the first house is the neighbor of the last one.
Meanwhile, the security system for these houses remain the same as
for those in the previous street.

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.

 /*************************************************************************
     > File Name: LeetCode213.c
     > Author: Juntaran
     > Mail: [e-mail]
     > Created Time: Wed 11 May 2016 17:11:02 PM CST
  ************************************************************************/

 /*************************************************************************

     House Robber II

     Note: This is an extension of House Robber.

     After robbing those houses on that street,
     the thief has found himself a new place for his thievery
     so that he will not get too much attention.
     This time, all houses at this place are arranged in a circle.
     That means the first house is the neighbor of the last one.
     Meanwhile, the security system for these houses remain the same as
     for those in the previous street.

     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.

  ************************************************************************/

 #include <stdio.h>

 /*
     跟198的区别在于现在是一个环形，首尾不能同时get
     所以分成两种情况：
     1：从头取到尾-1
     2：从头+1取到尾
     rob过程相同，然后比较两种情况大小
 */
 int rob( int* nums, int numsSize )
 {
     if( numsSize ==  )
     {
         return ;
     }
     if( numsSize ==  )
     {
         return nums[];
     }

     int max = ;
     int prev1 = ;
     int prev2 = ;

     int i, temp;

     temp  = ;
     prev1 = ;
     prev2 = ;
     for( i=; i<=numsSize-; i++ )
     {
         temp = prev1;
         prev1 = (prev2+nums[i])>prev1 ? (prev2+nums[i]) : prev1;
         prev2 = temp;
     }
     max = prev1;

     temp  = ;
     prev1 = ;
     prev2 = ;
     for( i=; i<=numsSize-; i++ )
     {
         temp = prev1;
         prev1 = (prev2+nums[i])>prev1 ? (prev2+nums[i]) : prev1;
         prev2 = temp;
     }
     max = max>prev1 ? max : prev1;

     return max;
 }

 int main()
 {
     int nums[] = { ,,,,,, };
     int numsSize = ;

     int ret = rob( nums, numsSize );
     printf("%d\n", ret);
     return ;
 }