Paint House II 解答

Question

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?

Solution

这里时间复杂度是O(nk)，说明要求我们用O(k)的时间计算每一层的新的cost。

cost'[i] = costs[m][i] + min{cost[0], cost[1], ..., cost[i - 1], cost[i + 1], ..., cost[k - 1]}

原想法是对每一个i重新计算，时间复杂度是O(k²)。包含了大量的重复计算

其实我们只需求出cost[]序列的最小值和第二小的值。Time complexity O(k)

public class Solution {
    public int minCostII(int[][] costs) {
        if (costs == null || costs.length < 1) {
            return 0;
        }
        int m = costs.length, k = costs[0].length;
        int[] cost = new int[k];
        int[] tmp = new int[k];
        for (int i = 0; i < k; i++) {
            cost[i] = costs[m - 1][i];
        }
        for (int i = m - 2; i >= 0; i--) {
            // calculate most and second minimum number
            int[] min = calcMin(cost);
            for (int j = 0; j < k; j++) {
                // if cost[j] is minimum, then add second minimum with costs[i][j]
                if (cost[j] == min[0]) {
                    cost[j] = min[1] + costs[i][j];
                } else {
                    // if cost[j] is not minimum, then add minimum with costs[i][j]
                    cost[j] = min[0] + costs[i][j];
                }
            }
        }
        if (k < 2) {
            return cost[0];
        }
        int[] result = calcMin(cost);
        return result[0];
    }
    
    private int[] calcMin(int[] nums) {
        if (nums == null || nums.length < 2) {
            return new int[0];
        }
        int[] mins = new int[2];
        mins[0] = Math.min(nums[0], nums[1]);
        mins[1] = Math.max(nums[0], nums[1]);
        for (int i = 2; i < nums.length; i++) {
            if (nums[i] < mins[0]) {
                mins[1] = mins[0];
                mins[0] = nums[i];
            } else if (nums[i] < mins[1]) {
                mins[1] = nums[i];
            }
        }
        return mins;
    }
}