原题传送门

1. 题目描述

2. Solution 1

1、思路分析
Dynamic Programming
Intuition: At every i-th house we have two choices to make, i.e., rob the i-th house or don't rob it.

Case1 : Don't rob the i-th house - then we can rob the i-1 th house...so we will have max money robbed till i-1 th house.
Case 2 : Rob the i-th house - then we cann't rob the i-1 th house but we can rob i-2 th house....so we will have max money robbed till i-2 th house + money of i-th house.

Example:
1.) If the array is [1,5,3] then robber will rob the 1st index house because arr[1] > arr[0]+arr[2] (i.e., at
last index, arr[i-1] > arr[i-2]+arr[i])
2.) If the array is [1,2,3] then robber will rob the 0th and 2nd index house because arr[0]+arr[2] > arr[1] (i.e
., at last index, arr[i-2] + arr[i] > arr[i-1])

1》状态定义: dp[i] 表示偷第i户人家后获得的最大收益。
2》边界: dp[0] = nums[0], dp[1] = max{nums[0], nums[1]};
3》状态转移方程:
dp[i] = max{nums[i] + dp[i-2], dp[i-1]};
case 2 case 1

2、代码实现

package Q0199.Q0198HouseRobber;

/*
    Dynamic Programming
    Intuition: At every i-th house we have two choices to make, i.e., rob the i-th house or don't rob it.

    Case1 : Don't rob the i-th house - then we can rob the i-1 th house...so we will have max money robbed till i-1
    th house
    Case 2 : Rob the i-th house - then we cann't rob the i-1 th house but we can rob i-2 th house....so we will have
    max money robbed till i-2 th house + money of i-th house.
    Example:
    1.) If the array is [1,5,3] then robber will rob the 1st index house because arr[1] > arr[0]+arr[2] (i.e., at
    last index, arr[i-1] > arr[i-2]+arr[i])
    2.) If the array is [1,2,3] then robber will rob the 0th and 2nd index house because arr[0]+arr[2] > arr[1] (i.e
    ., at last index, arr[i-2] + arr[i] > arr[i-1])
 */
public class Solution {
    public int rob(int[] nums) {
        if (nums.length == 1) return nums[0];
        // 1. 状态定义，i-th house的收益
        int[] dp = new int[nums.length];
        // 2. 初始状态
        dp[0] = nums[0];
        dp[1] = Math.max(nums[0], nums[1]);
        for (int i = 2; i < nums.length; i++)
            // 3. 状态转移方程    Case 2              Case 1
            dp[i] = Math.max(nums[i] + dp[i - 2], dp[i - 1]);
        return dp[nums.length - 1];
    }
}

3、复杂度分析
时间复杂度: O(n)
空间复杂度: O(n)

3. Solution 2

1、思路分析
降低空间复杂度。用变量代替数组。

2、代码实现

package Q0199.Q0198HouseRobber;

public class Solution2 {
    public int rob(int[] nums) {
		/*
		 思路: Dynamic Programming
		 设f(k)=从前k个house中抢到的最大金额,A[i]=第i个house中的金额
		 n = 1, f(1) = A[1]
		 n = 2, f(2) = max(A[1], A[2])
		 n = 3, f(3) = A[3] + A[1], if rob the third house
		  		or f(3) = A[2] , if don't rob the third house
		 ==> transition equation: f(k) = max( (f(k-2) + A[k]), f(k-1))
		 Time complexity: O(n), where n is the length of nums.
		 Space complexity: O(1)
		 */
        int prevMax = 0;
        int currMax = 0;
        for (int x : nums) {
            int temp = currMax;
            currMax = Math.max(prevMax + x, currMax);
            prevMax = temp;
        }
        return currMax;
    }
}

3、复杂度分析
时间复杂度: O(n)
空间复杂度: O(1)

标签：arr,nums,House,rob,0198,th,house,LeetCode,dp
来源： https://www.cnblogs.com/junstat/p/16329088.html