标签：follow ace paint font [1] 代码 runtime 没有 for
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?
此题是paint house的升级，和paint house不同之处在于，每个房子粉刷的颜色不再是常数了，而是k，此题想了很久没有什么思路，便直接看大神的解题思路了。
首先定义两个最小值，一个是第一小的值，一个是第二小的值，代表遍历到每个房子各个颜色的成本最小值，和第二小的值。在粉刷下一个房子的时候，如果之前的房子所包含的颜色不在最小成本值里面，就把最小成本值加上去，如果在最小成本值里面，则把第二小的值加进去。代码如下：
public class Solution {
public int minCostII(int[][] costs) {
if(costs==null||costs.length==0||costs[0].length==0) return 0;
int min1 = -1;
int min2 = -1;
int m = costs.length;
int k = costs[0].length;
for(int i=0;i<m;i++){
int last1= min1;
int last2 = min2;
min1 = -1;
min2 = -1;
for(int j=0;j<k;j++){
if(last1!=j){
costs[i][j]+=last1<0?0:costs[i-1][last1];
}else{
costs[i][j]+=last1<0?0:costs[i-1][last2];
}
if(min1<0||costs[i][j]<costs[i][min1]){
min2 = min1;
min1 = j;
}else if(min2<0||costs[i][j]<costs[i][min2]){
min2 = j;
}
}
}
return costs[m-1][min1];
}
}
标签：follow ace paint font [1] 代码 runtime 没有 for
原文地址：http://www.cnblogs.com/codeskiller/p/6357885.html
