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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?
解法一：O(nkk) ...
public class Solution { public int minCostII(int[][] costs) { if(costs==null||costs.length==0){ return 0; } for(int i=1; i<costs.length; i++) { for(int j=0; j<costs[0].length;j++) { int min = Integer.MAX_VALUE; for(int k=0; k<costs[0].length;k++) { if(k==j) continue; min = Math.min(min,costs[i-1][k]); } costs[i][j] += min; } } int n = costs.length-1; int min = Integer.MAX_VALUE; for(int j=0; j<costs[0].length;j++) { min = Math.min(min,costs[n][j]); } return min; } }
解法二：O(nk)
Explanation: dp[i][j] represents the min paint cost from house 0 to house i when house i use color j; The formula will be dp[i][j] = Math.min(any k!= j| dp[i-1][k]) + costs[i][j].
Take a closer look at the formula, we don‘t need an array to represent dp[i][j], we only need to know the min cost to the previous house of any color and if the color j is used on previous house to get prev min cost, use the second min cost that are not using color j on the previous house. So I have three variable to record: prevMin, prevMinColor, prevSecondMin. and the above formula will be translated into: dp[currentHouse][currentColor] = (currentColor == prevMinColor? prevSecondMin: prevMin) + costs[currentHouse][currentColor].
public class Solution { public int minCostII(int[][] costs) { if(costs == null || costs.length == 0 || costs[0].length == 0) return 0; int n = costs.length, k = costs[0].length; if(k == 1) return (n==1? costs[0][0] : -1); int prevMin = 0, prevMinInd = -1, prevSecMin = 0;//prevSecMin always >= prevMin for(int i = 0; i<n; i++) { int min = Integer.MAX_VALUE, minInd = -1, secMin = Integer.MAX_VALUE; for(int j = 0; j<k; j++) { int val = costs[i][j] + (j == prevMinInd? prevSecMin : prevMin); if(minInd< 0) {min = val; minInd = j;}//when min isn‘t initialized else if(val < min) {//when val < min, secMin = min; min = val; minInd = j; } else if(val < secMin) { //when min<=val< secMin secMin = val; } } prevMin = min; prevMinInd = minInd; prevSecMin = secMin; } return prevMin; } }
reference:https://leetcode.com/discuss/60625/fast-dp-java-solution-runtime-o-nk-space-o-1
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原文地址：http://www.cnblogs.com/hygeia/p/5126089.html
