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])