先用樸素的動態規劃，狀態轉移方程
ans[i][j] = min(ans[i-1][j]+cost[i][k], ans[i][j])
89%時提示超時。

class Solution:
    # @param {int[][]} costs n x k cost matrix
    # @return {int} an integer, the minimum cost to paint all houses
    def minCostII(self, costs):
        # Write your code here
        if not costs:
            return
 
 0
        n = len(costs)
        m = len(costs[0])
        ans = [[9999999 for i in range(m)] for i in range(2)]
        ans[0] = costs[0]
        for i in range(1, n):
            for j in range(m):
                for k in range(m):
                    if j != k:
                        ans[1][j] = min(ans[0
 
][k] + costs[i][j], ans[1][j])           
            ans[0] = ans[1]
            ans[1] = [9999999 for i in range(m)]       
        return min(ans[0])