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.
Notice
All costs are positive integers.
Example
Given costs = [[14,2,11],[11,14,5],[14,3,10]] return 10
house 0 is blue, house 1 is green, house 2 is blue, 2 + 5 + 3 = 10
LeetCode上的原题，请参见我之前的博客Paint House。
解法一：
class Solution {
public:
/**
* @param costs n x 3 cost matrix
* @return an integer, the minimum cost to paint all houses
*/
int minCost(vector<vector<int>>& costs) {
if (costs.empty() || costs[].empty()) return ;
vector<vector<int>> dp = costs;
for (int i = ; i < dp.size(); ++i) {
for (int j = ; j < ; ++j) {
dp[i][j] += min(dp[i - ][(j + ) % ], dp[i - ][(j + ) % ]);
}
}
return min(dp.back()[], min(dp.back()[], dp.back()[]));
}
};
解法二：
class Solution {
public:
/**
* @param costs n x 3 cost matrix
* @return an integer, the minimum cost to paint all houses
*/
int minCost(vector<vector<int>>& costs) {
if (costs.empty() || costs[].empty()) return ;
vector<vector<int>> dp = costs;
for (int i = ; i < dp.size(); ++i) {
dp[i][] += min(dp[i - ][], dp[i - ][]);
dp[i][] += min(dp[i - ][], dp[i - ][]);
dp[i][] += min(dp[i - ][], dp[i - ][]);
}
return min(dp.back()[], min(dp.back()[], dp.back()[]));
}
};