265. Paint House II 房子涂色K种选择的版本
[抄题]:
There are a row of n houses, each house can be painted with one of the k colors. 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 k
cost matrix. For example, costs[0][0]
is the cost of painting house 0 with color 0; costs[1][2]
is the cost of painting house 1 with color 2, and so on... Find the minimum cost to paint all houses.
Note:
All costs are positive integers.
Example:
Input: [[1,5,3],[2,9,4]]
Output: 5
Explanation: Paint house 0 into color 0, paint house 1 into color 2. Minimum cost: 1 + 4 = 5;
Or paint house 0 into color 2, paint house 1 into color 0. Minimum cost: 3 + 2 = 5.
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
k个颜色就不知道怎么办了:还是试啊 再套一层循环 一个个加
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[一句话思路]:
三重循环, s 和 j相等的时候就continue掉
[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]:
- i j是主变量,所以cost[i][j]都得用, dp[i][j]数组在不变的情况下就是它自己本身
dp[i][j] = Math.min(dp[i][j], dp[i - 1][s] + costs[i][j]);
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
所以cost[i][j]都得用, dp[i][j]数组在不变的情况下就是它自己本身
[复杂度]:Time complexity: O(n*k*k) Space complexity: O(n*k)
[算法思想:迭代/递归/分治/贪心]:
贪心
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
[是否头一次写此类driver funcion的代码] :
class Solution {
public int minCostII(int[][] costs) {
//cc
if (costs == null || costs.length == 0) return 0; //ini: dp[][], dp[0][k]
int n = costs.length, k = costs[0].length;
int[][] dp = new int[n][k];
for (int j = 0; j < k; j++) {
dp[0][j] = costs[0][j];
} //for loop: continue;
for (int i = 1; i < n; i++) {
for (int j = 0; j < k; j++) {
dp[i][j] = Integer.MAX_VALUE;
for (int s = 0; s < k; s++) {
if (s == j) continue;
dp[i][j] = Math.min(dp[i][j], dp[i - 1][s] + costs[i][j]);
}
}
} //return: compare each costs[i][k]
int res = Integer.MAX_VALUE;
for (int j = 0; j < k; j++) {
res = Math.min(res, dp[n - 1][j]);
} return res;
}
}
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