[Leetcode] Paint House I & II

Paint House

There is a row of n houses, each house can is painted with one of the three colors:red, blue or green. The cost of painting a certain color is different. You had to paint all the houses such, that no, and adjacent houses have the same color.

The cost of painting a certain color was represented by a cost n x 3 matrix. For example, was the cost of costs[0][0] painting House 0 with color red; costs[1][2] the cost of Painting House 1 with color green, And so on ... Find the minimum cost-to-paint all houses.

Note:
All costs is positive integers.

1 classSolution {2  Public:3     intMincost (vector<vector<int>>&costs) {4         if(Costs.empty ())return 0;5          for(inti =1; I < costs.size (); ++i) {6              for(intj =0; J <3; ++j) {7COSTS[I][J] + = min (costs[i-1[(j+1)%3], costs[i-1[(j+2)%3]);8             }9         }Ten         returnMin (Costs.back () [0], Min (Costs.back () [1], Costs.back () [2])); One     } A};

Paint House II

There is a row of n houses, each house can is painted with one of the K colors. The cost of painting a certain color is different. You had to paint all the houses such, that no, and adjacent houses have the same color.

The cost of painting a certain color was represented by a cost n x k matrix. For example, was the cost of costs[0][0] painting House 0 with color 0; costs[1][2] is the cost of Painting House 1 with color 2, ... Find the minimum cost-to-paint all houses.

Note:
All costs is positive integers.

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

1 classSolution {2  Public:3     intMincostii (vector<vector<int>>&costs) {4         if(Costs.empty () | | costs[0].empty ())return 0;5         intn = costs.size (), k = costs[0].size ();6vector<int>Min1 (k), min2 (k);7          for(inti =1; I < costs.size (); ++i) {8min1[0] =Int_max;9              for(intj =1; J < K; ++j) {TenMin1[j] = min (min1[j-1], costs[i-1][j-1]); One             } Amin2[k-1] =Int_max; -              for(intj = k-2; J >=0; --j) { -Min2[j] = min (min2[j+1], costs[i-1][j+1]); the             } -              for(intj =0; J < K; ++j) { -COSTS[I][J] + =min (min1[j], min2[j]); -             } +         } -         intres =Int_max; +          for(Auto C:costs.back ()) { Ares =min (res, c); at         } -         returnRes; -     } -};

Quickly find the minimum method of removing an element in an array: Define two arrays, min1[i] and Min2[i] record the minimum interval from left to right to point I and right to left to the first I bit, then the minimum value of the first bit of the minus is the Min (Min1[i], min2[i]).

[Leetcode] Paint House I & II