Paint House(medium) 房子涂色

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, costs0 is the cost of painting house 0 with color red; costs1 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.

解题思路

dp[i][j] -- the min cost for house i on painting color j

dp[i][R] = cost[i][R] + Math.min(dp[i - 1][B], dp[i - 1][G]);

dp[i][B] = cost[i][B] + Math.min(dp[i - 1][R], dp[i -1 ][G]);

dp[i][G] = cost[i][G] + Math.min(dp[i - 1][R], dp[i - 1][B]);

Final staus: min(dp[n - 1][R], dp[n - 1][B], dp[n - 1][G]);

public int minCost(int[][] costs) {
    if (costs == null || costs.length == 0) {
        return 0;
    }

    int[][] dp = new int[costs.length][3];

    dp[0][0] = costs[0][0];
    dp[0][1] = costs[0][1];
    dp[0][2] = costs[0][2];

    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]));
}