LeetCode-Paint House II

标签：

There are a row of n houses, each house can be painted with one of the k colors. 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 k cost matrix. For example, costs[0][0] is the cost of painting house 0 with color 0; costs[1][2] is the cost of painting house 1 with color 2, and so on... Find the minimum cost to paint all houses.

Note:
All costs are positive integers.

Follow up:
Could you solve it in O(nk) runtime?

Analysis:

dp[i][j]: the min cost of painting the first i houses with the last house painting color j.

dp[i][j] = max(dp[i-1][l] | l!=j) + costs[i][j];

We only need to maintain the first min cost and the second min cost of the last house.

Solution:

public class Solution {
    public int minCostII(int[][] costs) {
        if (costs.length == 0 || costs[0].length == 0)
            return 0;

        int n = costs.length, k = costs[0].length;
        int[] dp = new int[k];
        int lastMin1 = 0, lastMin2 = 0;
        for (int i = 0; i < n; i++) {
            int curMin1 = Integer.MAX_VALUE - 1, curMin2 = Integer.MAX_VALUE;
            for (int j = 0; j < k; j++) {
                dp[j] = ((dp[j] == lastMin1) ? lastMin2 : lastMin1) + costs[i][j];
                if (dp[j] <= curMin1) {
                    curMin2 = curMin1;
                    curMin1 = dp[j];
                } else if (dp[j] < curMin2) {
                    curMin2 = dp[j];
                }
            }
            lastMin1 = curMin1;
            lastMin2 = curMin2;
        }
        return lastMin1;
    }
}

LeetCode-Paint House II

标签：

原文地址：http://www.cnblogs.com/lishiblog/p/5836176.html