256. Paint House

问题描述

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.

题目链接：

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思路分析

有n座房子需要被涂上红绿蓝三种颜色，每座房子涂三种颜色的cost存在一个二维数组中，每座相邻的房子颜色不能相同，求最小的总cost。

动态规划题目，先判断数组是否为空。每座房子有三种颜色可以选择，所以有三种初始条件。然后开始循环，下一座房子选择的颜色为上一房子选择的两种颜色的最小值加这座房子颜色的cost。一直遍历到最后，然后返回三者中最小的就可以了。

代码

递归

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    int minCost(vector<vector<int>>& costs) {
        if (costs.size() == 0)
            return 0;
        int lastR = costs[0][0];
        int lastB = costs[0][1];
        int lastG = costs[0][2];
        for (int i = 1; i < costs.size(); i++){
            int curR = min(lastB, lastG) + costs[i][0];
            int curB = min(lastR, lastG) + costs[i][1];
            int curG = min(lastR, lastB) + costs[i][2];

            lastR = curR;
            lastB = curB;
            lastG = curG;
        }
        return min(lastR, min(lastB, lastG));
    }
};

时间复杂度：$O(n)$
空间复杂度：$O(1)$

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反思

动态规划题目，现在有些不会做了，这是一个有三种情况的，需要注意。