LeetCode 256. Paint House（房子涂色）_房子涂颜色动态规划-优快云博客

本文介绍如何使用动态规划解决最小涂房子成本问题，包括三种不同的实现方式：直接方法、通用形式和内存优化方法。通过遍历每个房子并计算最小成本，最终找到所有房子的最小总成本。

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原题网址：https://leetcode.com/problems/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, costs[0][0] is the cost of painting house 0 with color red; costs[1][2] 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.

方法：动态规划：

public class Solution {
    public int minCost(int[][] costs) {
        int red = 0, blue = 0, green = 0;
        for(int i=0; i<costs.length; i++) {
            int nr = Math.min(blue, green) + costs[i][0];
            int nb = Math.min(red, green) + costs[i][1];
            int ng = Math.min(red, blue) + costs[i][2];
            red = nr;
            blue = nb;
            green = ng;
        }
        return Math.min(red, Math.min(blue, green));
    }
}

或者形式上更通用一点：

public class Solution {
    public int minCost(int[][] costs) {
        if (costs == null || costs.length == 0) return 0;
        int[] current = new int[3];
        int[] buf = new int[3];
        for(int c=0; c<3; c++) current[c] = costs[0][c];
        for(int h=1; h<costs.length; h++) {
            int[] prev = current;
            current = buf;
            current[0] = Math.min(prev[1], prev[2]) + costs[h][0];
            current[1] = Math.min(prev[0], prev[2]) + costs[h][1];
            current[2] = Math.min(prev[0], prev[1]) + costs[h][2];
            buf = prev;
        }
        int min = current[0];
        if (current[1] < min) min = current[1];
        if (current[2] < min) min = current[2];
        return min;
    }
}

也可以使用取模的方法来实现内存优化（九章算法老师提供的方法）：

public class Solution {
    public int minCost(int[][] costs) {
        int[][] total = new int[2][3];
        for(int i=1; i<=costs.length; i++) {
            total[i%2][0] = Math.min(total[(i-1)%2][1]+costs[i-1][0], total[(i-1)%2][2]+costs[i-1][0]);
            total[i%2][1] = Math.min(total[(i-1)%2][0]+costs[i-1][1], total[(i-1)%2][2]+costs[i-1][1]);
            total[i%2][2] = Math.min(total[(i-1)%2][0]+costs[i-1][2], total[(i-1)%2][1]+costs[i-1][2]);
        }
        return Math.min(total[costs.length%2][0], Math.min(total[costs.length%2][1], total[costs.length%2][2]));
    }
}