256. Paint House房屋染色

［抄题］：

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.

 [暴力解法]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

［思维问题］：

以为要用i , j的二维矩阵：就3种情况，枚举所有情况就行了

［一句话思路］：

求和取前一步最小值，从而实现步步最小

［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

1. 理解过程：把所有值都累加到nums[n]中

［二刷］：

［三刷］：

［四刷］：

［五刷］：

  [五分钟肉眼debug的结果]：

［总结］：

 求和类型的三步dp，套用坐标型模板

［复杂度］：Time complexity: O(n) Space complexity: O(n)

［英文数据结构或算法，为什么不用别的数据结构或算法］：

［关键模板化代码］：

标准格式i = 0，i < n，return f[n - 1]

for (int i = 1; i < n; i++) {
            costs[i][0] += Math.min(costs[i - 1][1], costs[i - 1][2]);
            costs[i][1] += Math.min(costs[i - 1][0], costs[i - 1][2]);
            costs[i][2] += Math.min(costs[i - 1][0], costs[i - 1][1]);
        }
        //answer
        return Math.min(costs[n - 1][0], Math.min(costs[n - 1][1],costs[n - 1][2]));

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

265. Paint House II 数学问题，感觉并没有什么联系

 [代码风格] ：

public class Solution {
    /**
     * @param costs: n x 3 cost matrix
     * @return: An integer, the minimum cost to paint all houses
     */
    public int minCost(int[][] costs) {
        //corner case
        if (costs.length == 0 || costs[0].length == 0) {
            return 0;
        }
        //get sum
        int n = costs.length;
        for (int i = 1; i < n; i++) {
            costs[i][0] += Math.min(costs[i - 1][1], costs[i - 1][2]);
            costs[i][1] += Math.min(costs[i - 1][0], costs[i - 1][2]);
            costs[i][2] += Math.min(costs[i - 1][0], costs[i - 1][1]);
        }
        //answer
        return Math.min(costs[n - 1][0], Math.min(costs[n - 1][1],costs[n - 1][2]));
    }
}