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[LintCode/LeetCode] Paint House I & Paint Hous

ChristmasBoy / 888人閱讀

摘要：在原數(shù)組上動規(guī)，每一行對應(yīng)一個(gè)房子，每一個(gè)元素代表從第一行的房子到這一行的房子選擇這一種顏色所花的最小開銷。所以每個(gè)元素該元素的值上一行兩個(gè)與該元素不同列元素的值的較小者。不過這次要記錄三個(gè)變量本行最小值，本行第二小值，本行最小值下標(biāo)。

Paint House Problem

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, costs0 is the cost of painting house 0 with color red; costs1 is the cost of painting house 1 with color green, and so on... Find the minimum cost to paint all houses.

Example

Given costs = [[14,2,11],[11,14,5],[14,3,10]] return 10

house 0 is blue, house 1 is green, house 2 is blue, 2 + 5 + 3 = 10

Note

在原數(shù)組上動規(guī)，每一行對應(yīng)一個(gè)房子，每一個(gè)元素代表從第一行的房子到這一行的房子選擇這一種顏色所花的最小開銷。所以每個(gè)元素 = 該元素的值 + 上一行兩個(gè)與該元素不同列元素的值的較小者。
動規(guī)到

Solution 
public class Solution {
    public int minCost(int[][] cost) {
        if (cost == null || cost.length == 0) return 0;
        int n = cost.length - 1;
        for (int i = 1; i <= n; i++) {
            cost[i][0] += Math.min(cost[i-1][1], cost[i-1][2]);
            cost[i][1] += Math.min(cost[i-1][0], cost[i-1][2]);
            cost[i][2] += Math.min(cost[i-1][0], cost[i-1][1]);
        }
        return Math.min(cost[n][0], Math.min(cost[n][1], cost[n][2]));
    }
}
Paint House II Problem

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, costs0 is the cost of painting house 0 with color 0; costs1 is the cost of painting house 1 with color 2, and so on... Find the minimum cost to paint all houses.

Example

Given n = 3, k = 3, costs = [[14,2,11],[11,14,5],[14,3,10]] return 10

house 0 is color 2, house 1 is color 3, house 2 is color 2, 2 + 5 + 3 = 10

Challenge

Could you solve it in O(nk)?

Note

和Paint House的思路一樣，依然是在cost數(shù)組上更新最小花銷之和。不過這次要記錄三個(gè)變量：本行最小值，本行第二小值，本行最小值下標(biāo)。這三個(gè)變量每行更新一次，同時(shí)也要存儲上一次循環(huán)的三個(gè)變量。這樣做的目的是當(dāng)前一行最小值和之前一行最小值在同一列時(shí)，取之前一行的次小值，也就避免了相鄰兩行取同種顏色的問題。

Solution 
public class Solution {
    public int minCostII(int[][] cost) {
        if (cost == null || cost.length == 0) return 0;
        int preMin = 0, preSec = 0, preIndex = -1;
        int n = cost.length, k = cost[0].length;
        for (int i = 0; i < n; i++) {
            int curMin = Integer.MAX_VALUE, curSec = Integer.MAX_VALUE, curIndex = 0;
            for (int j = 0; j < k; j++) {
                cost[i][j] += preIndex == j ? preSec : preMin;
                if (cost[i][j] < curMin) {
                    curSec = curMin;
                    curMin = cost[i][j];
                    curIndex = j;
                }
                else if (cost[i][j] < curSec) {
                    curSec = cost[i][j];
                }
            }
            preMin = curMin;
            preSec = curSec;
            preIndex = curIndex;
        }
        return preMin;
    }
}

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