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  • 256. Paint House

    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.

    dynamic programming

    在costs的基础上,每一行每个元素累加与上一次不同颜色的cost,在最后一行取最小值

    time: O(m), space: O(1)  -- m:  # rows of costs

    class Solution {
        public int minCost(int[][] costs) {
            if(costs == null || costs.length == 0 || costs[0].length == 0) return 0;
            
            for(int i = 1; i < costs.length; i++) {
                costs[i][0] = costs[i][0] + Math.min(costs[i-1][1], costs[i-1][2]);
                costs[i][1] = costs[i][1] + Math.min(costs[i-1][0], costs[i-1][2]);
                costs[i][2] = costs[i][2] + Math.min(costs[i-1][0], costs[i-1][1]);
            }
    
            int n = costs.length - 1;
            return Math.min(costs[n][0], Math.min(costs[n][1], costs[n][2]));
        }
    }
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  • 原文地址:https://www.cnblogs.com/fatttcat/p/10130211.html
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