• [LC] 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.

    class Solution {
        public int minCost(int[][] costs) {
            if (costs == null || costs.length == 0 || costs[0].length == 0) {
                return 0;
            }
            int curRed = costs[0][0];
            int curBlue = costs[0][1];
            int curGreen = costs[0][2];
            int prevRed = curRed;
            int prevBlue = curBlue;
            int prevGreen = curGreen;
            for (int i = 1; i < costs.length; i++) {
                curRed = Math.min(prevBlue, prevGreen) + costs[i][0];
                curBlue = Math.min(prevRed, prevGreen) + costs[i][1];
                curGreen = Math.min(prevRed, prevBlue) + costs[i][2];
                prevRed = curRed;
                prevBlue = curBlue;
                prevGreen = curGreen;
            }
            return Math.min(curRed, Math.min(curBlue, curGreen));
        }
    }
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  • 原文地址:https://www.cnblogs.com/xuanlu/p/12053778.html
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