515. Paint House

最新推荐文章于 2021-05-07 09:32:01 发布

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最新推荐文章于 2021-05-07 09:32:01 发布

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原文链接：http://www.cnblogs.com/lawrenceSeattle/p/10272975.html

本文详细解析了PaintHouse问题的算法解决方案，旨在找到给定房屋行中涂色的最低成本，确保相邻房屋颜色不同。通过动态规划方法，计算出所有房屋涂色的最低总成本。

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515. 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.

Example
Given costs = [[14,2,11],[11,14,5],[14,3,10]] return 10

house 0 is blue, house 1 is green, house 2 is blue, 2 + 5 + 3 = 10

Notice
All costs are positive integers.

public class Solution {
    /**
     * @param costs: n x 3 cost matrix
     * @return: An integer, the minimum cost to paint all houses
     */
    public int minCost(int[][] costs) {
        // 有三种颜色，相邻的两个不能是一样的颜色，求最低的成本
        if(costs == null || costs.length <= 0) return 0;
        int[][] dp = new int[costs.length][3];
        
        for(int i = 0; i< 3; i++) {
            dp[0][i] = costs[0][i];
        }
        for(int i = 1; i < costs.length; i++) {
            dp[i][0] = costs[i][0] + Math.min(dp[i-1][1], dp[i-1][2]);
            dp[i][1] = costs[i][1] + Math.min(dp[i-1][0], dp[i-1][2]);
            dp[i][2] = costs[i][2] + Math.min(dp[i-1][0], dp[i-1][1]);
        }
        return Math.min(dp[costs.length-1][0], Math.min(dp[costs.length-1][1], dp[costs.length-1][2]));
    }
}