Leetcode 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.

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解题思路：

Dynamic Programming

dp[i][j] = costs[i][j] + min(dp[i-1][(j+1)%3], dp[i-1][(j+2)%3])

需要维持dp[i] 和dp[i-1] 。

方法一：可以直接建立数组dp[costs.length][3]，花更多空间。

方法二：也可以建立两个preRow[3] 储存dp[i-1], currORw 储存dp[i], 节省空间，运行时间也短一些。

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Java code:

方法一：

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

方法二：

public class Solution {
    public int minCost(int[][] costs) {
        if(costs == null || costs.length == 0){
            return 0;
        }
        int[] preRow = costs[0];
        int len = costs.length;
        for(int i = 1; i < len; i++) {
            int[] currRow = new int[3];
            for(int j = 0; j < 3; j++){
                currRow[j] = costs[i][j] + Math.min(preRow[(j+1)%3], preRow[(j+2)%3]);
            }
            preRow = currRow;
        }
        return Math.min(Math.min(preRow[0], preRow[1]), preRow[2]);
    }
}

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Reference:

1. https://leetcode.com/discuss/63863/simple-15-line-code-with-o-n-time-and-o-1-memory-solution-java