[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]));
}
}

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