Paint House II

here 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?

Tips:

这道题是之前那道Paint House的拓展，那道题只让用红绿蓝三种颜色来粉刷房子，而这道题让我们用k种颜色，这道题不能用之前那题的解法，会TLE。

这题的解法的思路还是用DP，但是在找不同颜色的最小值不是遍历所有不同颜色，而是用min1和min2来记录之前房子的最小和第二小的花费的颜色，如果当前房子颜色和min1相同，那么我们用min2对应的值计算，反之我们用min1对应的值，这种解法实际上也包含了求次小值的方法，感觉也是一种很棒的解题思路

Code:

public class Solution {
    public int minCostII(int[][] costs) {
        if (costs == null || costs.length == 0 || costs[0] == null || costs[0].length == 0) {
            return 0;
        }
        int[] store = new int[costs[0].length];
        int pre1 = 0, pre2 = 0;
        int min1 = 0, min2 = 0;
        for (int i = 0; i < costs.length; i++) {
            pre1 = min1;
            pre2 = min2;
            min1 = Integer.MAX_VALUE;
            min2 = Integer.MAX_VALUE;
            for (int j = 0; j < costs[0].length; j++) {
                if (store[j] != pre1 || pre1 == pre2) {
                    store[j] = costs[i][j] + pre1;
                }
                else  store[j] = costs[i][j] + pre2;
                if (store[j] <= min1) {
                    min2 = min1;
                    min1 = store[j];
                }
                else if (store[j] <= min2) {
                    min2 = store[j];
                }
            }
        }
        return min1;
    }
}