198.House Rubber

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Dynamic Programming， Easy

Question

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:

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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:

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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.

Answer

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class Solution {
    public int rob(int[] nums) {
        //Dynamic programming
        if(nums.length == 0)
            return 0;
        if(nums.length == 1)
            return nums[0];
        int[] v = new int[nums.length];
        v[0] = nums[0];
        v[1] = Math.max(nums[0], nums[1]);
        
        for(int i=2; i<nums.length; i++)
            v[i] = Math.max(v[i-2] + nums[i], v[i-1]);
        return v[nums.length-1];
    }
}

Running time: O(n)

Space: O(n)

Better Answer

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public int rob(int[] num) {
    int[][] dp = new int[num.length + 1][2];
    for (int i = 1; i <= num.length; i++) {
        dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1]);
        dp[i][1] = num[i - 1] + dp[i - 1][0];
    }
    return Math.max(dp[num.length][0], dp[num.length][1]);
}

dp[i][1] means we rob the current house and dp[i][0] means we don’t,

so it is easy to convert this to O(1) space

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public int rob(int[] num) {
    int rob = 0; //max monney can get if rob current house
    int notrob = 0; //max money can get if not rob current house
    for(int i=0; i<num.length; i++) {
        int currob = notrob + num[i]; //if rob current value, previous house must not be robbed
        notrob = Math.max(notrob, rob); //if not rob ith house, take the max value of robbed (i-1)th house and not rob (i-1)th house
        rob = currob;
    }
    return Math.max(rob, notrob);
}