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.
Follow up:
Could you solve it in O(nk) runtime?
这道题是之前那道 Paint House 的拓展，那道题只让用红绿蓝三种颜色来粉刷房子，而这道题让用k种颜色，这道题不能用之前那题的解法，会 TLE。这题的解法的思路还是用 DP，但是在找不同颜色的最小值不是遍历所有不同颜色，而是用 min1 和 min2 来记录之前房子的最小和第二小的花费的颜色，如果当前房子颜色和 min1 相同，那么用 min2 对应的值计算，反之用 min1 对应的值，这种解法实际上也包含了求次小值的方法，感觉也是一种很棒的解题思路，参见代码如下：
解法一：
class Solution {
public:
int minCostII(vector<vector<int>>& costs) {
if (costs.empty() || costs[].empty()) return ;
vector<vector<int>> dp = costs;
int min1 = -, min2 = -;
for (int i = ; i < dp.size(); ++i) {
int last1 = min1, last2 = min2;
min1 = -; min2 = -;
for (int j = ; j < dp[i].size(); ++j) {
if (j != last1) {
dp[i][j] += last1 < ? : dp[i - ][last1];
} else {
dp[i][j] += last2 < ? : dp[i - ][last2];
}
if (min1 < || dp[i][j] < dp[i][min1]) {
min2 = min1; min1 = j;
} else if (min2 < || dp[i][j] < dp[i][min2]) {
min2 = j;
}
}
}
return dp.back()[min1];
}
};
下面这种解法不需要建立二维 dp 数组，直接用三个变量就可以保存需要的信息即可，参见代码如下：
解法二：
class Solution {
public:
int minCostII(vector<vector<int>>& costs) {
if (costs.empty() || costs[].empty()) return ;
int min1 = , min2 = , idx1 = -;
for (int i = ; i < costs.size(); ++i) {
int m1 = INT_MAX, m2 = m1, id1 = -;
for (int j = ; j < costs[i].size(); ++j) {
int cost = costs[i][j] + (j == idx1 ? min2 : min1);
if (cost < m1) {
m2 = m1; m1 = cost; id1 = j;
} else if (cost < m2) {
m2 = cost;
}
}
min1 = m1; min2 = m2; idx1 = id1;
}
return min1;
}
};
Github 同步地址：
https://github.com/grandyang/leetcode/issues/265
类似题目：
参考资料：
https://leetcode.com/problems/paint-house-ii/
https://leetcode.com/problems/paint-house-ii/discuss/69509/Easiest-O(1)-space-JAVA-solution
https://leetcode.com/problems/paint-house-ii/discuss/69492/AC-Java-solution-without-extra-space