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?
[分析]
使用color[j]刷house[i]时我们希望知道刷house[0,i-1]且house[i-1]刷的不是color[j]时的最小cost,于是求解过程中我们需要保留的就是dp[i][j],表示刷完house[0, i]的最小cost且house[i]使用color[j],递推式就是dp[i][j] = min(dp[i - 1][0]...dp[i-1][k-1]) + costs[i][j]
思路1:直接实现上面的递推式,复杂度是O(n*k*k),非最优。
思路2:受Product of Array Except Self启发,使用两个额外数组:
minPrior[i][j]表示使用color[0,j]刷house[i]时刷house[0,i]需要的最小cost,
minAfter[i][j]表示使用color[0,j]刷house[i]时刷house[i,n-1]需要的最小cost.
则递推式变为dp[i][j] = min(minPrior[i-1][j-1], minAfter[i-1][j+1]) + costs[i][j]
复杂度是O(n*k)。因为变量较多且需要处理边界元素,很快bug free难度较高。
思路3:每次迭代house i前先计算出dp[i-1][]的最小值和次小值,
则递推式简化为dp[i][j]=(dp[i-1][j] == preMin ? preSecondMin : preMin) + costs[i][j]
如果dp[i-1][j]正好是刷前面所有house的最小值,由于相邻house不能同色,那这次我们就要用次小值(运气好,可能和最小值相等~) 这个思路很清晰容易实现,且空间复杂度还低,
谢谢yb君分享的巧妙方案~
public class Solution { public int minCostII(int[][] costs) { if (costs == null || costs.length == 0) return 0; int n = costs.length, k = costs[0].length; if (n > 1 && k == 1) { return 0; // ask interviewer whether return 0 or Integer.MAX } //dp[i][j]: minmum cost painting house 0-i with house i painted with color j int[][] dp = new int[n][k]; for (int j = 0; j < k; j++) dp[0][j] = costs[0][j]; for (int i = 1; i < n; i++) { int preMin = Integer.MAX_VALUE; int preSecondMin = Integer.MAX_VALUE; for (int j = 0; j < k; j++) { if (dp[i - 1][j] <= preMin) { preSecondMin = preMin; preMin = dp[i - 1][j]; } else if (dp[i - 1][j] < preSecondMin) { preSecondMin = dp[i - 1][j]; } } for (int j = 0; j < k; j++) { if (dp[i - 1][j] == preMin) { dp[i][j] = preSecondMin + costs[i][j]; } else { dp[i][j] = preMin + costs[i][j]; } } } int min = Integer.MAX_VALUE; for (int j = 0; j < k; j++) { min = Math.min(min, dp[n - 1][j]); } return min; } public int minCostII_Method1(int[][] costs) { if (costs == null || costs.length == 0) return 0; int n = costs.length, k = costs[0].length; if (n > 1 && k == 1) { return 0; } if (n == 1) { int min = costs[0][0]; for (int j = 1; j < k; j++) { min = Math.min(min, costs[0][j]); } return min; } int[][] dp = new int[n][k]; int[][] minPrior = new int[n][k]; int[][] minAfter = new int[n][k]; dp[0][0] = costs[0][0]; minPrior[0][0] = dp[0][0]; for (int j = 1; j < k; j++) { dp[0][j] = costs[0][j]; minPrior[0][j] = Math.min(minPrior[0][j - 1], dp[0][j]); } minAfter[0][k - 1] = dp[0][k - 1]; for (int j = k - 2; j >= 0; j--) { minAfter[0][j] = Math.min(minAfter[0][j + 1], dp[0][j]); } for (int i = 1; i < n; i++) { dp[i][0] = minAfter[i - 1][1] + costs[i][0]; minPrior[i][0] = dp[i][0]; for (int j = 1; j < k - 1; j++) { dp[i][j] = Math.min(minPrior[i - 1][j - 1], minAfter[i - 1][j + 1]) + costs[i][j]; minPrior[i][j] = Math.min(minPrior[i][j - 1], dp[i][j]); } dp[i][k - 1] = minPrior[i- 1][k - 2] + costs[i][k - 1];// at first minPrior[i][k - 2], make me crazy, cyb kill the bug minPrior[i][k - 1] = Math.min(minPrior[i][k - 2], dp[i][k - 1]); minAfter[i][k - 1] = dp[i][k - 1]; for (int j = k - 2; j >= 0; j--) { minAfter[i][j] = Math.min(minAfter[i][j + 1], dp[i][j]); } } return minPrior[n - 1][k - 1]; } }