LeetCode 265. Paint House II（房子涂色）

原题网址：https://leetcode.com/problems/paint-house-ii/

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?

方法：动态规划，求出最低成本和次低成本，如果房子的颜色和上一个房子最低成本的颜色相同，则涂上次低成本的颜色，否则涂最低成本的颜色。

这道题不需要我们分别求出与每一个颜色不同的最低成本值，只需要求出最低和次低，就可以有效避免颜色间隔的问题，这是关键。

public class Solution {
    private void find(int[] cost, int[] min, int[] color) {
        color[0] = -1;
        color[1] = -1;
        for(int i=0; i<cost.length; i++) {
            if (color[0] == -1 || cost[i] < min[0]) {
                min[0] = cost[i];
                color[0] = i;
            }
        }
        for(int i=0; i<cost.length; i++) {
            if (i != color[0] && (color[1] == -1 || cost[i] < min[1])) {
                min[1] = cost[i];
                color[1] = i;
            }
        }
    }
    public int minCostII(int[][] costs) {
        if (costs == null || costs.length == 0) return 0;
        int[] minCost = new int[2];
        int[] color = new int[2];
        int[] nextMinCost = new int[2];
        int[] nextColor = new int[2];
        find(costs[0], minCost, color);
        for(int i=1; i<costs.length; i++) {
            for(int j=0; j<costs[i].length; j++) {
                costs[i][j] += j!=color[0] ? minCost[0] : minCost[1];
            }
            find(costs[i], nextMinCost, nextColor);
            int[] t1 = minCost;
            minCost = nextMinCost;
            nextMinCost = t1;
            int[] t2 = color;
            color = nextColor;
            nextColor = t2;
        }
        return minCost[0];
    }
}

可以进一步优化得更简洁：

public class Solution {
    // 求出最低成本和次低成本
    private void mins(int[] cost, int[] min) {
        for(int i=0; i<cost.length; i++) {
            if (i==0 || cost[i] < cost[min[0]]) min[0] = i;
        }
        if (min[0] == 0) min[1] = 1; else min[1] = 0;
        for(int i=0; i<cost.length; i++) {
            if (i != min[0] && cost[i] < cost[min[1]]) min[1] = i;
        }
    }
    
    public int minCostII(int[][] costs) {
        if (costs == null || costs.length == 0) return 0;
        int[] minCost = new int[costs[0].length];
        int[] min = new int[2];
        for(int c=0; c<costs[0].length; c++) {
            minCost[c] = costs[0][c];
        }
        mins(minCost, min);
        for(int h=1; h<costs.length; h++) {
            int[] prevCost = minCost;
            minCost = new int[costs[0].length];
            for(int c=0; c<costs[h].length; c++) {
                if (c != min[0]) minCost[c] = prevCost[min[0]] + costs[h][c];
                else minCost[c] = prevCost[min[1]] + costs[h][c];
            }
            mins(minCost, min);
        }
        return minCost[min[0]];
    }
}

如果不分成两个步骤，即先计算每种颜色的最低成本，在寻找最低成本的两种颜色，就会比较罗嗦：

public class Solution {
    public int minCostII(int[][] costs) {
        if (costs == null || costs.length == 0 || costs[0].length == 0) return 0;
        int n = costs.length;
        int k = costs[0].length;
        int[][] mins = new int[n][2];

        int[][] colors = new int[n][2];
        for(int j=0; j<k; j++) {
            if (mins[0][0] == 0 || costs[0][j] < mins[0][0]) {
                colors[0][0] = j;
                mins[0][0] = costs[0][j];
            }
        }
        for(int j=0; j<k; j++) {
            if (j==colors[0][0]) continue;
            if (mins[0][1] == 0 || costs[0][j] < mins[0][1]) {
                colors[0][1] = j;
                mins[0][1] = costs[0][j];
            }
        }
        for(int i=1; i<n; i++) {
            // generate minimum
            for(int j=0; j<k; j++) {
                if (j!=colors[i-1][0]) {
                    if (mins[i][0] == 0 || mins[i-1][0] + costs[i][j] < mins[i][0]) {
                        colors[i][0] = j;
                        mins[i][0] = mins[i-1][0] + costs[i][j];
                    }
                } else {
                    if (mins[i][0] == 0 || mins[i-1][1] + costs[i][j] < mins[i][0]) {
                        colors[i][0] = j;
                        mins[i][0] = mins[i-1][1] + costs[i][j];
                    }
                }
                
            }
            // generate second minimum
            for(int j=0; j<k; j++) {
                if (j==colors[i][0]) continue;
                if (j!=colors[i-1][0]) {
                    if (mins[i][1] == 0 || mins[i-1][0] + costs[i][j] < mins[i][1]) {
                        colors[i][1] = j;
                        mins[i][1] = mins[i-1][0] + costs[i][j];
                    }
                } else {
                    if (mins[i][1] == 0 || mins[i-1][1] + costs[i][j] < mins[i][1]) {
                        colors[i][1] = j;
                        mins[i][1] = mins[i-1][1] + costs[i][j];
                    }
                }
            }
        }
        return mins[n-1][0];
    }
}