[Leetcode] 256. Paint House 解题报告

题目：

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.

思路：

算是一道简单的动态规划题目。我们定义red[i]表示将第i个房屋刷成红色时，前i个房屋的最小代价，则状态转移方程为red[i] = min(green[i - 1], blue[i - 1]) + costs[i][0]。green[i]和blue[i]的定义以及递推公式类似。最终返回red[size - 1], green[size - 1]和blue[size - 1]的最小值即可。

代码：

class Solution {public:    int minCost(vector<vector<int>>& costs) {        int size = costs.size();        if (size == 0) {            return 0;        }        vector<int> red(size, INT_MAX);        vector<int> green(size, INT_MAX);        vector<int> blue(size, INT_MAX);        red[0] = costs[0][0];        green[0] = costs[0][1];        blue[0] = costs[0][2];        for (int i = 1; i < size; ++i) {            red[i] = min(blue[i - 1], green[i - 1]) + costs[i][0];            green[i] = min(red[i - 1], blue[i - 1]) + costs[i][1];            blue[i] = min(red[i - 1], green[i - 1]) + costs[i][2];        }        return min(red[size - 1], min(green[size - 1], blue[size - 1]));    }};