难度: Easy
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.
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])