[LintCode] Paint House 粉刷房子

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()[]));
    }
};