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256. Paint House

难度: Easy

刷题内容

原题连接

内容描述

There are a row of n houses, each house can be painted with one of the three colors: red, blue or green. 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 3 cost matrix. For example, costs[0][0] is the cost of painting house 0 with color red; costs[1][2] is the cost of painting house 1 with color green, and so on... Find the minimum cost to paint all houses.

Note:
All costs are positive integers.

Example:

Input: [[17,2,17],[16,16,5],[14,3,19]]
Output: 10
Explanation: Paint house 0 into blue, paint house 1 into green, paint house 2 into blue. 
             Minimum cost: 2 + 5 + 3 = 10.

解题方案

思路 1 - 时间复杂度: O(N)- 空间复杂度: O(N)******

dp[i][j]代表第i个house用第j种颜色,前i个house的总cost

beats 99.68%

class Solution:
    def minCost(self, costs):
        """
        :type costs: List[List[int]]
        :rtype: int
        """
        if not costs or len(costs) == 0:
            return 0
        
        dp = [costs[0][:] for i in range(len(costs))]
        for i in range(1, len(costs)):
            dp[i][0] = min(dp[i-1][1], dp[i-1][2]) + costs[i][0]
            dp[i][1] = min(dp[i-1][0], dp[i-1][2]) + costs[i][1]
            dp[i][2] = min(dp[i-1][0], dp[i-1][1]) + costs[i][2]
        return min(dp[-1])