198. House Robber [LeetCode]

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本文探讨了一个专业窃贼如何在不触动相邻房屋警报系统的情况下，通过动态规划算法来决定抢劫哪些房屋以获得最大收益的问题。文章通过两个实例说明了算法的具体应用过程。

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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.

Example 1:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
             Total amount you can rob = 2 + 9 + 1 = 12.

//动态规划超简单

int rob(int* nums, int numsSize) {
    if (numsSize == 0) return 0;
    if (numsSize == 1) return nums[0];
    if (numsSize == 2) return fmax(nums[0], nums[1]);
    
    int *f = (int *)calloc(numsSize, sizeof(int));
    f[0] = nums[0];
    f[1] = fmax(nums[0], nums[1]);
    for (int i = 2; i < numsSize; i++) {
        f[i] = fmax(nums[i] + f[i-2], f[i-1]);
    }
    int ret = f[numsSize - 1];
    free(f);
    return ret;
}