[LeetCode] Paint House I & II

Paint House

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.

 class Solution {
 public:
     int minCost(vector<vector<int>>& costs) {
         if (costs.empty()) return ;
         for (int i = ; i < costs.size(); ++i) {
             for (int j = ; j < ; ++j) {
                 costs[i][j] += min(costs[i-][(j+)%], costs[i-][(j+)%]);
             }
         }
         return min(costs.back()[], min(costs.back()[], costs.back()[]));
     }
 };

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?

 class Solution {
 public:
     int minCostII(vector<vector<int>>& costs) {
         if (costs.empty() || costs[].empty()) return ;
         int n = costs.size(), k = costs[].size();
         vector<int> min1(k), min2(k);
         for (int i = ; i < costs.size(); ++i) {
             min1[] = INT_MAX;
             for (int j = ; j < k; ++j) {
                 min1[j] = min(min1[j-], costs[i-][j-]);
             }
             min2[k-] = INT_MAX;
             for (int j = k - ; j >= ; --j) {
                 min2[j] = min(min2[j+], costs[i-][j+]);
             }
             for (int j = ; j < k; ++j) {
                 costs[i][j] += min(min1[j], min2[j]);
             }
         }
         int res = INT_MAX;
         for (auto c : costs.back()) {
             res = min(res, c);
         }
         return res;
     }
 };

快速找到数组中去掉某个元素的最小值方法：定义两个数组，min1[i]与min2[i]分别记录从左向右到第i位与从右向左到第i位的区间最小值，那么去掉第i位的最小值就是min(min1[i], min2[i])。