LeetCode – Paint House (Java)
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.
Java Solution
A typical DP problem.
public int minCost(int[][] costs) { if(costs==null||costs.length==0) return 0; for(int i=1; i<costs.length; i++){ costs[i][0] += Math.min(costs[i-1][1], costs[i-1][2]); costs[i][1] += Math.min(costs[i-1][0], costs[i-1][2]); costs[i][2] += Math.min(costs[i-1][0], costs[i-1][1]); } int m = costs.length-1; return Math.min(Math.min(costs[m][0], costs[m][1]), costs[m][2]); } |
Or a different way of writing the code without original array value changed.
public int minCost(int[][] costs) { if(costs==null||costs.length==0){ return 0; } int[][] matrix = new int[3][costs.length]; for(int j=0; j<costs.length; j++){ if(j==0){ matrix[0][j]=costs[j][0]; matrix[1][j]=costs[j][1]; matrix[2][j]=costs[j][2]; }else{ matrix[0][j]=Math.min(matrix[1][j-1], matrix[2][j-1])+costs[j][0]; matrix[1][j]=Math.min(matrix[0][j-1], matrix[2][j-1])+costs[j][1]; matrix[2][j]=Math.min(matrix[0][j-1], matrix[1][j-1])+costs[j][2]; } } int result = Math.min(matrix[0][costs.length-1], matrix[1][costs.length-1]); result = Math.min(result, matrix[2][costs.length-1]); return result; } |
<pre><code> String foo = "bar"; </code></pre>
-
Sarthak Sota