*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, are the cost of costs[0][0] painting House 0 with color 0; costs[1][2] are the cost of Painting House 1 with color 2, and S O on ... Find the minimum cost-to-paint all houses.

Note:
All costs is positive integers.

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

Solution One: O (NKK) ...

 Public classSolution { Public intMincostii (int[] costs) {    if(costs==NULL|| Costs.length==0){        return0; }         for(intI=1; i<costs.length; i++)    {         for(intj=0; j<costs[0].length;j++)        {            intMin =Integer.max_value;  for(intk=0; k<costs[0].length;k++)            {                if(K==J)Continue; Min= Math.min (min,costs[i-1][k]); } Costs[i][j]+=min; }    }        intn = costs.length-1; intMin =Integer.max_value;  for(intj=0; j<costs[0].length;j++) {min=math.min (Min,costs[n][j]); }    returnmin; }}

Solution Two: O (NK)

Explanation:dp[i][j] represents the min paint cost from House 0 to house I if House I use color j; The formula is 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 was used on previous house to get Prev min cost, with the second min cost tha T is a using color J on the previous house. So I had three variable to record:prevmin, Prevmincolor, Prevsecondmin. And the above formula would be translated into:dp[currenthouse][currentcolor] = (CurrentColor = = Prevmincolor? prevsecondm In:prevmin) + Costs[currenthouse][currentcolor].

 Public classSolution { Public intMincostii (int[] costs) {    if(Costs = =NULL|| Costs.length = = 0 | | Costs[0].length = = 0)return0; intn = costs.length, k = costs[0].length; if(k = = 1)return(N==1 costs[0][0]: 1); intPrevmin = 0, Prevminind =-1, prevsecmin = 0;//prevsecmin always >= prevmin     for(inti = 0; i<n; i++) {        intmin = integer.max_value, Minind =-1, secmin =Integer.max_value;  for(intj = 0;  j<k; J + +) {            intval = 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< secminSecmin =Val; }} prevmin=min; Prevminind=Minind; Prevsecmin=secmin; }    returnprevmin;}}

Reference:https://leetcode.com/discuss/60625/fast-dp-java-solution-runtime-o-nk-space-o-1

*paint House II