[LeetCode265]Paint House II

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.

Follow up:
Could you solve it in O(nk) runtime?

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这题这么不简单，我当然不会。参考了大家的O(nk)方法。

dp[j]: 刷前i个房子且最后一个房子用的是j颜色的minimum cost。

min1: 刷前i个房子最后用某个颜色时最小的cost。

min2: 刷前i个房子最后用另外某个颜色时第二小的cost。

对于当前房子，如果之前的房子的minimum cost并不是min1 我们可以直接选取min1的颜色，否则选取min2的颜色。因为只要相邻两个房子颜色不一样即可。

class Solution {
public:
int minCostII(vector<vector<int>>& costs) {
if (costs.empty()) return 0;
int n = costs.size(), k = costs[0].size(), min1 = 0, min2 = 0;
vector<int> dp(k,0);
for(int i = 0; i<n; ++i){
int tmp1 = min1, tmp2 = min2;
min1 = min2 = INT_MAX;
for(int j = 0; j<k; ++j){
dp[j] = ((dp[j] == tmp1) ? tmp2 : tmp1) + costs[i][j];
if(min1 <= dp[j]) min2 = min(min2, dp[j]);
else{
min2 = min1;
min1 = dp[j];
}
}
}
return min1;
}
};