[Leetcode] Paint House 房子涂色

Paint House

There are a row of n houses, each house can be painted with one of the three colors: red, blue or green. The cost of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent houses have the same color.

The cost of painting each house with a certain color is represented by a n x 3 cost matrix. For example, costs0 is the cost of painting house 0 with color red; costs1 is the cost of painting house 1 with color green, and so on... Find the minimum cost to paint all houses.

Note: All costs are positive integers.

动态规划

复杂度

时间 O(N) 空间 O(1)

思路

直到房子i，其最小的涂色开销是直到房子i-1的最小涂色开销，加上房子i本身的涂色开销。但是房子i的涂色方式需要根据房子i-1的涂色方式来确定，所以我们对房子i-1要记录涂三种颜色分别不同的开销，这样房子i在涂色的时候，我们就知道三种颜色各自的最小开销是多少了。我们在原数组上修改，可以做到不用空间。

代码

public class Solution {
    public int minCost(int[][] costs) {
        if(costs != null && costs.length == 0) return 0;
        for(int i = 1; i < costs.length; i++){
            // 涂第一种颜色的话，上一个房子就不能涂第一种颜色，这样我们要在上一个房子的第二和第三个颜色的最小开销中找最小的那个加上
            costs[i][0] = costs[i][0] + Math.min(costs[i - 1][1], costs[i - 1][2]);
            // 涂第二或者第三种颜色同理
            costs[i][1] = costs[i][1] + Math.min(costs[i - 1][0], costs[i - 1][2]);
            costs[i][2] = costs[i][2] + Math.min(costs[i - 1][0], costs[i - 1][1]);
        }
        // 返回涂三种颜色中开销最小的那个
        return Math.min(costs[costs.length - 1][0], Math.min(costs[costs.length - 1][1], costs[costs.length - 1][2]));
    }
}

Paint House II

There are a row of n houses, each house can be painted with one of the k colors. The cost of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent houses have the same color.

The cost of painting each house with a certain color is represented by a n x k cost matrix. For example, costs0 is the cost of painting house 0 with color 0; costs1 is the cost of painting house 1 with color 2, and so on... Find the minimum cost to paint all houses.

Note: All costs are positive integers.

Follow up: Could you solve it in O(nk) runtime?

动态规划

复杂度

时间 O(N) 空间 O(1)

思路

和I的思路一样，不过这里我们有K个颜色，不能简单的用Math.min方法了。如果遍历一遍颜色数组就找出除了自身外最小的颜色呢？我们只要把最小和次小的都记录下来就行了，这样如果和最小的是一个颜色，就加上次小的开销，反之，则加上最小的开销。

代码

public class Solution {
    public int minCostII(int[][] costs) {
        if(costs != null && costs.length == 0) return 0;
        int prevMin = 0, prevSec = 0, prevIdx = -1;
        for(int i = 0; i < costs.length; i++){
            int currMin = Integer.MAX_VALUE, currSec = Integer.MAX_VALUE, currIdx = -1;
            for(int j = 0; j < costs[0].length; j++){
                costs[i][j] = costs[i][j] + (prevIdx == j ? prevSec : prevMin);
                // 找出最小和次小的，最小的要记录下标，方便下一轮判断
                if(costs[i][j] < currMin){
                    currSec = currMin;
                    currMin = costs[i][j];
                    currIdx = j;
                } else if (costs[i][j] < currSec){
                    currSec = costs[i][j];
                }
            }
            prevMin = currMin;
            prevSec = currSec;
            prevIdx = currIdx;
        }
        return prevMin;
    }
}