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.
- All costs are positive integers.
C Solution 1:
#define MIN(a, b) ((a) < (b) ? (a) : (b))
int minCost(int** costs, int costsRowSize, int costsColSize) {
if (costsRowSize < 1) return 0;
int i;
for (i = 1; i < costsRowSize; i++) {
costs[i][0] += MIN(costs[i - 1][1], costs[i - 1][2]);
costs[i][1] += MIN(costs[i - 1][0], costs[i - 1][2]);
costs[i][2] += MIN(costs[i - 1][0], costs[i - 1][1]);
}
return MIN(MIN(costs[i - 1][0], costs[i - 1][1]), costs[i - 1][2]);
}
C Solution 2:
#define MIN(a, b) ((a) < (b) ? (a) : (b))
int minCost(int** costs, int costsRowSize, int costsColSize) {
int i, A = 0, B = 0, C = 0;
for (i = 0; i < costsRowSize; i++) {
int _A = costs[i][0] + MIN(B, C);
int _B = costs[i][1] + MIN(A, C);
int _C = costs[i][2] + MIN(A, B);
A = _A, B = _B, C = _C;
}
return MIN(MIN(A, B), C);
}
Python Solution 1:
class Solution(object):
def minCost(self, costs):
"""
:type costs: List[List[int]]
:rtype: int
"""
if not costs: return 0
for i in range(1, len(costs)):
costs[i][0] += min(costs[i - 1][1], costs[i - 1][2])
costs[i][1] += min(costs[i - 1][0], costs[i - 1][2])
costs[i][2] += min(costs[i - 1][0], costs[i - 1][1])
return min(costs[len(costs) - 1])
Python Solution 2:
class Solution(object):
def minCost(self, costs):
"""
:type costs: List[List[int]]
:rtype: int
"""
return min(reduce(lambda (A, B, C), (a, b, c): (a + min(B, C), b + min(A, C), c + min(A, B)), costs, [0] * 3))
Summary:
- Solution and Good Solution.
LeetCode: 256. Paint House
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