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?
解题思路
和I的思路一样，不过这里我们有K个颜色，不能简单的用Math.min方法了。如果遍历一遍颜色数组就找出除了自身外最小的颜色呢？我们只要把最小和次小的都记录下来就行了，这样如果和最小的是一个颜色，就加上次小的开销，反之，则加上最小的开销。
public int minCostII(int[][] costs) {
if(costs != null && costs.length == 0) return 0;
int prevMin = 0, prevSec = 0, prevIdx = -1;
for(int i = 0; i < costs.length; i++){
int currMin = Integer.MAX_VALUE, currSec = Integer.MAX_VALUE, currIdx = -1;
for(int j = 0; j < costs[0].length; j++){
costs[i][j] = costs[i][j] + (prevIdx == j ? prevSec : prevMin);
// 找出最小和次小的，最小的要记录下标，方便下一轮判断
if(costs[i][j] < currMin){
currSec = currMin;
currMin = costs[i][j];
currIdx = j;
} else if (costs[i][j] < currSec){
currSec = costs[i][j];
}
}
prevMin = currMin;
prevSec = currSec;
prevIdx = currIdx;
}
return prevMin;
}