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[Leetcode]Paint House II

2015-09-15 17:28 549 查看
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?
/*algorithm: dp soluton ,similar paint house problem
for path[0..i]
level i; dp[i][0] = min(dp[i-1][1],...,dp[i-1][k-1]) + dp[i][0];
dp[i][1] = min(dp[i-1][0],dp[i-1][2],...,dp[i-1][k-1]) + dp[i][1]
dp[i][j] = min(dp[i-1][0],...dp[i-1][j-1],dp[i-1][j+1],..dp[i-1][k-1]) + dp[i][j]
the path min cost is min(dp[n-1][0...k-1])
time O(n*k*k) space O(1)
*/
int minCostII(vector<vector<int> >&costs) {
int n = costs.size();
if(n < 1)return 0;
int k = costs[0].size();
for(int i = 1;i < n;i++){
for(int j = 0;j < k;j++){
int cost = INT_MAX;
for(int m = 0;m < k;m++){
if(m != j)cost = min(costs[i-1][m],cost);
}
costs[i][j] += cost;
}
}

//for last level
int cost = INT_MAX;
for(int i = 0;i < k;i++)
cost = min(cost,cost[n-1][i]);
return cost;
}
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