house robber - DP

198. House Robber
213. House Robber II
337. House Robber III
256. Paint House
265. Paint House II
276. Paint Fence

198. House Robber

  • 一个array,每个值代表该house的value,不能偷相邻的house,最多偷多少
  • every house has two possible states, rob/not rob
  • if we don't rob the current house, the value we can rob is max(rob-pre-house, don't-rob-pre-house)
  • if we rob the current house (we cannot rob the previous house), the value is
    (don't-rob-pre-house + the value in the current house)
class Solution {
    public int rob(int[] nums) {
        int n = nums.length;
        int[][] dp = new int[n + 1][2];  //two states, 0 is not rob, 1 is rob
        for (int i = 1; i <= n; i++) {
            dp[i][0] = Math.max(dp[i-1][0], dp[i-1][1]);
            dp[i][1] = dp[i-1][0] + nums[i-1];
        }
        return Math.max(dp[n][0], dp[n][1]);
    }
}

213. House Robber II

  • 这些房子构成环,第一个房子和最后一个房子相邻
  • 两种方式,第1个到n-1个,第2个到第n个
class Solution {
    //house[1] and house[n] are adjacent
    //so we can rob from house[1] to house[n-1];
    //or rob from house[2] to house[n]
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        if (nums.length == 1) return nums[0];
        int n = nums.length;
        int first = robFromTo(nums, 1, n-1);
        int second = robFromTo(nums, 2, n);
        return Math.max(first, second);
    }
    
    private int robFromTo(int[] nums, int from, int to) {
        int n = nums.length;
        int[][] dp = new int[n+1][2];
        for (int i = from; i <= to; i++) {
            dp[i][0] = Math.max(dp[i-1][0], dp[i-1][1]);
            dp[i][1] = dp[i-1][0] + nums[i-1];
        }
        return Math.max(dp[to][0], dp[to][1]);
    }
}

337. House Robber III

  • 所有的房子构成一个binary tree,相邻的两个房子不能都偷
  • up down 的方式,调用很多次dfs
  • bottom up 利用返回的int[]存储信息
    a good explanation
//这种方法调用过多dfs,我一直在想有没有memorization的方法,记住某个node选和不选时最多能rob的数
class Solution {
    int sum = 0;
    public int rob(TreeNode root) {
        if (root == null) return 0;
        int first = dfs(root, false);
        int second = dfs(root, true);
        return Math.max(first, second);
    }
    private int dfs(TreeNode root, boolean select) {
        if (root == null) return 0;
        int sum = 0;
        if (select) {
            sum += root.val;
            //自己选了,child就不能选
            sum += dfs(root.left, false);
            sum += dfs(root.right, false);
        } else {
            //自己没选,child可选可不选
            sum += Math.max(dfs(root.left, false), dfs(root.left, true));
            sum += Math.max(dfs(root.right, false), dfs(root.right, true));
        }
        return sum;
    }
}
//当从一个node返回时,要携带什么信息?
//how much i can rob if i select this node, and
//how much i can rob if i don't select this node
class Solution {
    public int rob(TreeNode root) {
        int[] res = dfs(root);     
        return Math.max(res[0], res[1]);
    }
    private int[] dfs(TreeNode root) {
        if (root == null) {
            return new int[2];
        }
        //[0] not choose the node, [1]choose the node
        int[] left = dfs(root.left);
        int[] right = dfs(root.right);
        int[] cur = new int[2];
        //if we do not choose the current node, we can either choose or not the left, right
        cur[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);  
        //if we choose the current node, we cannot choose the left, right
        cur[1] = left[0] + right[0] + root.val;
        return cur;
    }
}

256. Paint House

  • 与House Robber非常像,不允许相邻house被染成同样的颜色
  • paint每个房子每种颜色的价格都不一样,找最小需要多少钱将其全染完
  • key point,2D Araay, record the minimum cost with house i and color j
class Solution {
    public int minCost(int[][] costs) {
        int n = costs.length;
        int[][] dp = new int[n+1][3];
        for (int i = 1; i <= n; i++) {
            dp[i][0] = Math.min(dp[i-1][1], dp[i-1][2]) + costs[i-1][0];
            dp[i][1] = Math.min(dp[i-1][0], dp[i-1][2]) + costs[i-1][1];
            dp[i][2] = Math.min(dp[i-1][0], dp[i-1][1]) + costs[i-1][2];
        }
        return Math.min(Math.min(dp[n][0], dp[n][1]), dp[n][2]);
    }
}

265. Paint House II

  • Paint House的提高版,可以用O(nk2),其中k2是k个现在房子color和k个前一个房子color的组合
  • 其实只需要最低和第二低的前一个房子的color, 这样的是O(nk)
class Solution {
    //O(kn)
    //two pinters point to the corresponding color of the lowest cost 
    //and the correspongding color of second lowest cost
    public int minCostII(int[][] costs) {
        if (costs == null || costs.length == 0) return 0;
        int n = costs.length, k = costs[0].length;
        int[] pre = new int[k];
        int[] cur = new int[k];
        int min1 = -1, min2 = -1;
        for (int j = 0; j < k; j++) {
            pre[j] = costs[0][j];
            if (min1 == -1 || pre[j] < pre[min1]) {
                min2 = min1; min1 = j;
            } else if (min2 == -1 || pre[j] < pre[min2]) {
                min2 = j;
            }
        }
        for (int i = 1; i < n; i++) {
            int last1 = min1, last2 = min2;
            min1 = -1; min2 = -1;
            for (int j = 0; j < k; j++) {
                if (j != last1) {
                    cur[j] = pre[last1] + costs[i][j];
                } else {
                    cur[j] = pre[last2] + costs[i][j];
                }
                //update min1, min2
                if (min1 == -1 || cur[j] < cur[min1]) {
                    min2 = min1; min1 = j;
                } else if (min2 == -1 || cur[j] < cur[min2]) {
                    min2 = j;
                }
            }
            //update pre
            for (int j = 0; j < k; j++) {
                pre[j] = cur[j];
            }
        }
        return pre[min1];
    }
}

276. Paint Fence

  • 不能有连续3个相同的颜色
  • 第 i 和 i-1 post颜色不同时所有的可能性
  • 第 i 和 i-1 post颜色相同时所有的可能性
class Solution {
    public int numWays(int n, int k) {
        if (n == 0) return 0;
        if (n == 1) return k;
        int[] dp = new int[n+1];
        dp[0] = 0;
        dp[1] = k;
        dp[2] = k * k;
        for (int i = 3; i <= n; i++) {
            dp[i] += dp[i-1] * (k-1);  //第i和第i-1个不同时有多少情况
            dp[i] += dp[i-2] * (k-1);  //第i和第i-1相同,so 第i-2必须和第i-1不同
        }
        return dp[n];
    }
}

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