leetcode265- Paint House II- hard

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?

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. (出处：http://www.cnblogs.com/grandyang/p/5322870.html)

细节：DP初始化后之后遍历都是从i = 1开始算的，不要重新算0，会错

实现：

class Solution {
    public int minCostII(int[][] costs) {
        if (costs == null || costs.length == 0 || costs[0].length == 0) {
            return 0;
        }
        
        int[][] dp = new int[costs.length][costs[0].length];
        int preMin = Integer.MAX_VALUE;
        int preMinIdx = -1;
        int preMin2 = Integer.MAX_VALUE;
        
        for (int j = 0; j < costs[0].length; j++) {
            dp[0][j] = costs[0][j];
            if (dp[0][j] < preMin) {
                preMin2 = preMin;
                preMin = dp[0][j];
                preMinIdx = j;
            } else if (dp[0][j] < preMin2){
                preMin2 = dp[0][j];                
            }
        }   
             
        for (int i = 1; i < costs.length; i++) {
            int newPreMin = Integer.MAX_VALUE;
            int newPreMin2 = Integer.MAX_VALUE;
            int newPreMinIdx = -1;
            for(int j = 0; j < costs[0].length; j++) {
                if (j == preMinIdx) {
                    dp[i][j] = costs[i][j] + preMin2;
                } else {
                    dp[i][j] = costs[i][j] + preMin;
                }
                if (dp[i][j] < newPreMin) {
                    newPreMin2 = newPreMin;
                    newPreMin = dp[i][j];
                    newPreMinIdx = j;
                } else if (dp[i][j] < newPreMin2) {
                    newPreMin2 = dp[i][j];
                }
            }
            preMin = newPreMin;
            preMin2 = newPreMin2;
            preMinIdx = newPreMinIdx;        
        }
        
        int result = Integer.MAX_VALUE;
        for (int j = 0; j < dp[0].length; j++) {
            result = Math.min(result, dp[dp.length - 1][j]);
        }
        return result;
    }
}