[LeetCode] 256. Paint House 粉刷房子

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

Note:
All costs are positive integers.

 

解题思路：

房子i的最小涂色开销是房子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。

State: dp[i][j] // three colors: j = 0 or 1 or 2, 

Function:

    dp[i][0] = dp[i][0] + min(dp[i - 1][1], dp[i -1][2])

    dp[i][1] = dp[i][1] + min(dp[i - 1][0], dp[i - 1][2])

    dp[i][2] = dp[i][2] + min(dp[i - 1][0], dp[i - 1][1])　　　　

Initialize: dp = costs

Return: min(dp[n][0], dp[n][1], dp[n][2])

java 1: Time: O(n), Space: O(n)

public class Solution {
    public int minCost(int[][] costs) {
        int len = costs.length;
        if(costs != null && len == 0) return 0;
        int[][] dp = costs;
        for(int i = 1; i < len; i++){
            dp[i][0] = costs[i][0] + Math.min(costs[i - 1][1], costs[i - 1][2]);
            dp[i][1] = costs[i][1] + Math.min(costs[i - 1][0], costs[i - 1][2]);
            dp[i][2] = costs[i][2] + Math.min(costs[i - 1][0], costs[i - 1][1]);
        }
        return Math.min(dp[len - 1][0], Math.min(dp[len - 1][1], dp[len - 1][2]));
    }
  
    public static void main(String args[]) {
        int[][] costs = new int[][]{{14,2,11},{11,14,5},{14,3,10}};
        Solution sol = new Solution();
        System.out.println(sol.minCost(costs));
    }
}

　　　

Java 2: Time: O(n), Space: O(1)

class Solution {
    public int minCost(int[][] costs) {
        if(costs != null && costs.length == 0) return 0;        
        // 直接用原数组
        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]));
    }
}