You are a professional thief to steal street plan of the house. Each room are in possession of some cash, the only constraints affecting steal your neighbor's house is equipped with anti-theft system communicate with each other, if two adjacent houses on the same night, thieves broke into the system will automatically alarm .
Given a representative from each non-negative integer array of Housing storage amount calculated in case you do not touch the alarm device, can steal the maximum amount to.
Example 1:
Input: [1,2,3,1] Output: 4 Explanation: Housing theft No. 1 (amount = 1), then the theft Housing 3 (amount = 3).
The maximum amount to theft = 3 + 1 = 4. Example 2:Input: [2,7,9,3,1] Output: 12 Explanation: Housing theft No. 1 (amount = 2), 3 theft Housing (amount = 9), followed by theft. 5
No. Housing (amount = 1).
The maximum amount to theft = 2 + 9 + 1 = 12.Source: stay button (LeetCode) link: https: //leetcode-cn.com/problems/house-robber
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int max(int a, int b)
{
return a>b?a:b;
}
int rob(int* nums, int numsSize){
//动态规划类型，考虑每一个最大值都可以有哪些子问题构成
//例如：我要从前四个里面偷取，怎么偷呢！
//细思极恐，好几种可能。怎么办呢？记住，编程时就是将乱麻给捋顺
//这些可能可以分为以下几大类：肯定含有第一家的，肯定含有第二家的但不含第一家，肯定含第
//三家的但不含第一第二家，肯定含有第四家但不含有第一第二第三家的，从这些可能中找到最大的
//所以设dp[i]表示从i到n家的最大价值
//dp[i]=max(num[i]+dp[i+2],num[i+1]+dp[i+3].......)
if(numsSize==0)
{
return 0;
}
int* dp = (int*)calloc(numsSize,sizeof(int));
dp[numsSize-1]=nums[numsSize-1];
for(int i = numsSize-2;i>=0;i-- )
{
for(int j = i;j<=numsSize-1;j++)
{
dp[i]=max(dp[i],nums[j]+(j+2<numsSize?dp[j+2]:0));
}
}
return dp[0];
}
The results:
by
displaying details of
execution when used:
0 MS
, defeated in all C submission
100.00%
of user
memory consumption:
7.1 MB
, defeated in all C submission
26.94%
of users
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