Paint House 515

Question

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.

Notice

All costs are positive integers.

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

Solution

这道题因为只有3中颜色，所以很简单。dp[i][j]表示第i幢房子涂j的颜色最小的总和，即从前一幢房子的状态dp[i-1][] (k != j)中选一个最小的再加上给第i幢房子涂j颜色的cost。如果直接在costs上修改，则不用单独开dp的空间，可以优化空间。

时间复杂度O(n)，空间O(1)。

代码如下：

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) {
        // Write your code here
        if(costs == null || costs.length == 0 || costs[0].length == 0){
            return 0;
        }

        int n = costs.length;
        //直接在原数组上修改不用耗费额外空间
        for(int i = 1; i < n; i++){
            costs[i][0] = costs[i][0] + Math.min(costs[i - 1][1], costs[i - 1][2]);
            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]);
        }

        return Math.min(costs[n - 1][0], Math.min(costs[n - 1][1], costs[n - 1][2]));
    }
}