LeetCode 198 House Robber（基础DP）

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.
题目大意：你是一个专业的小偷，现在有一条街，街上的每一间房子里面都有一定价值的财物。每间房子之间都有警报器，如果连续两间房子被盗，那么警报器会自动报警。现在你要对这一条街上的房子行窃，问行窃的最大收益是多少。

解题思路：设dp[i]是行窃到第i间房子时的最大收益，则可得状态转移方程dp[i] = max(dp[i-2] + nums[i], dp[i-1])，根据状态转移方程可很方便得写出代码。

代码如下：

long long max(long long x, long long y){
return x > y ? x : y;
}

int rob(int* nums, int numsSize) {
if(numsSize == 0) return 0;
long long* dp = malloc(numsSize * sizeof *dp);
//dp[i]表示抢劫到第i个房子时的最大收益(下标从0开始)
dp[0] = nums[0];
dp[1] = max(dp[0], nums[1]);
for(int i = 2;i < numsSize;i++){
dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
}
return dp[numsSize - 1];
}