LeetCode OJ-198. House Robber

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.

题目要求是，不让警报响起并且盗得最多的金钱，相邻两个房子被盗则会引起警报。这里可以用动态规划解决，我们假定dp[i]表示到第i间房子时，所得到的最大收获，由此可得状态方程：

dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);

若将第i间房子的金钱盗走，则代表i - 1间房子未被盗，所以就与dp[i - 2]相加，结果与dp[i - 1](到第i - 1间房子时所得的最大收获)相比。这里需要清楚一点，dp[i - 1]是否包括第i - 1间房子的金钱并不重要，若包括，它比dp[i - 2] + nums[i]大，就将dp[i]赋值为dp[i - 1]代表第i间房子未被盗；若不包括，那它的值应与dp[i - 2]相同，无影响。具体代码如下：

#define max(m, n) (m > n ? m : n) 

int rob(int* nums, int numsSize) {
    int res = 0;
    int *dp = malloc(numsSize * sizeof(int));
    int i;
    memset(dp, 0, sizeof(dp));
    for (i = 0; i < numsSize; ++i) {
        //第1项和第2项需做特殊处理
        if (i - 1 < 0) {
            dp[i] = max(0 + nums[i], 0);
            continue;
        }
        if (i - 2 < 0) {
            dp[i] = max(0 + nums[i], dp[i - 1]);
            continue;
        }
        
        dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
    }
    
    res = dp[numsSize - 1];
    free(dp);
    return res;
}