[LeetCode] 256. Paint House_Easy tag: Dynamic Programming

Posted 2020-12-30

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.

思路为DP, 关系式为

#A[i][0] += min(A[i-1][1:])
#A[i][1] += min(A[i-1][0], A[i-1][2])
#A[i][2] += min(A[i-1][:2])

 

Code    T; O(n)     S: O(n)  可以利用inplace或者rolling array降为O(1)

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