LeetCode 256/265 Paint House系列问题

题目原文

256. Paint House

Difficulty: Easy

There are a row of n houses, each house can be painted with one of the three colors: red, blue or green. 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 _3_ cost matrix. For example, costs[0][0] is the cost of painting house 0 with color red; costs[1][2] is the cost of painting house 1 with color green, and so on… Find the minimum cost to paint all houses.

Note:
All costs are positive integers.

Example:

Input: [[17,2,17],[16,16,5],[14,3,19]]
Output: 10
Explanation: Paint house 0 into blue, paint house 1 into green, paint house 2 into blue. Minimum cost: 2 + 5 + 3 = 10.

解法

class Solution {
    public int minCost(int[][] costs) {
        if (costs == null || costs.length == 0) {
            return 0;
        }
        int m = costs.length;
        for (int i = 1; i < m; i++) {
            costs[i][0] += Math.min(costs[i - 1][1], costs[i - 1][2]);
            costs[i][1] += Math.min(costs[i - 1][0], costs[i - 1][2]);
            costs[i][2] += Math.min(costs[i - 1][0], costs[i - 1][1]);
        }
        return Math.min(costs[m - 1][0], Math.min(costs[m - 1][1], costs[m - 1][2]));
    }
}

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题目原文

265. Paint House II

Difficulty: Hard

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.

Example:

Input: [[1,5,3],[2,9,4]]
Output: 5
Explanation:
Paint house 0 into color 0, paint house 1 into color 2. Minimum cost: 1 + 4 = 5;
Or paint house 0 into color 2, paint house 1 into color 0. Minimum cost: 3 + 2 = 5.

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

解法

Language: Java

class Solution {
    public int minCostII(int[][] costs) {
        if (costs == null || costs.length == 0) {
            return 0;
        }
        int m = costs.length;
        int k = costs[0].length;
        // 上一列的最小值和次小值，当前列的最小值和次小值
        int[] min = {0, 0, Integer.MAX_VALUE, Integer.MAX_VALUE};
        // 上一列最小值坐标，当前列最小值坐标
        int[] minIndex = {-1, -1};
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < k; j++) {
                if (j == minIndex[0]) {
                    costs[i][j] += min[1];
                } else {
                    costs[i][j] += min[0];
                }
                if (costs[i][j] <= min[2]) {
                    min[3] = min[2];
                    min[2] = costs[i][j];
                    minIndex[1] = j;
                } else if (costs[i][j] < min[3]) {
                    min[3] = costs[i][j];
                }
            }
            min[0] = min[2];
            min[1] = min[3];
            min[2] = Integer.MAX_VALUE;
            min[3] = Integer.MAX_VALUE;
            minIndex[0] = minIndex[1];
        }
        return min[0];
    }
}