LeetCode 198.打家劫舍

链接：https://leetcode-cn.com/problems/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.

Example 1:

Input: nums = [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: nums = [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.

Constraints:

• 0 <= nums.length <= 100
• 0 <= nums[i] <= 400
  思路：动态规划

public int rob(int[] nums) {
    if (nums.length == 0) {
        return 0;
    }
    // 子问题：
    // f(k) = 偷 [0..k) 房间中的最大金额

    // f(0) = 0
    // f(1) = nums[0]
    // f(k) = max{ rob(k-1), nums[k-1] + rob(k-2) }

    int N = nums.length;
    int[] dp = new int[N+1];
    dp[0] = 0;
    dp[1] = nums[0];
    for (int k = 2; k <= N; k++) {
        dp[k] = Math.max(dp[k-1], nums[k-1] + dp[k-2]);
    }
    return dp[N];
}

作者：nettee
链接：https://leetcode-cn.com/problems/house-robber/solution/dong-tai-gui-hua-jie-ti-si-bu-zou-xiang-jie-cjavap/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

优化空间

public int rob(int[] nums) {
    int prev = 0;
    int curr = 0;

    // 每次循环，计算“偷到当前房子为止的最大金额”
    for (int i : nums) {
        // 循环开始时，curr 表示 dp[k-1]，prev 表示 dp[k-2]
        // dp[k] = max{ dp[k-1], dp[k-2] + i }
        int temp = Math.max(curr, prev + i);
        prev = curr;
        curr = temp;
        // 循环结束时，curr 表示 dp[k]，prev 表示 dp[k-1]
    }

    return curr;
}

作者：nettee
链接：https://leetcode-cn.com/problems/house-robber/solution/dong-tai-gui-hua-jie-ti-si-bu-zou-xiang-jie-cjavap/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。