zoukankan      html  css  js  c++  java
  • [LeetCode] 265. Paint House II 粉刷房子

    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.

    Note: All costs are positive integers.

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

    解题思路:

    这道题是Paint House的拓展,这题的解法的思路还是用DP,那道题只让用红绿蓝三种颜色来粉刷房子,而这道题让我们用k种颜色,这道题不能用之前那题的Math.min方法了,会TLE。只要把最小和次小的都记录下来就行了,用preMin和PreSec来记录之前房子的最小和第二小的花费的颜色,如果当前房子颜色和min1相同,那么我们用min2对应的值计算,反之我们用min1对应的值,这种解法实际上也包含了求次小值的方法。

    State: dp[i][j]

    Function: dp[i][j] = costs[i][j] + preMin or costs[i][j] + preSec

    Initialize: preMin = 0 , preSec = 0

    Return: dp[n][preMin]

    Java: Time: O(n), Space: O(1)

    public class Solution {
        public int minCostII(int[][] costs) {
            if(costs != null && costs.length == 0) return 0;
            int prevMin = 0, prevSec = 0, prevIdx = -1;
            for(int i = 0; i < costs.length; i++){
                int currMin = Integer.MAX_VALUE, currSec = Integer.MAX_VALUE, currIdx = -1;
                for(int j = 0; j < costs[0].length; j++){
                    costs[i][j] = costs[i][j] + (prevIdx == j ? prevSec : prevMin);
                    // 找出最小和次小的,最小的要记录下标,方便下一轮判断
                    if(costs[i][j] < currMin){
                        currSec = currMin;
                        currMin = costs[i][j];
                        currIdx = j;
                    } else if (costs[i][j] < currSec){
                        currSec = costs[i][j];
                    }
                }
                prevMin = currMin;
                prevSec = currSec;
                prevIdx = currIdx;
            }
            return prevMin;
        }
    }
    

      

    相似题目:

    256. Paint House

  • 相关阅读:
    Server Tomcat v8.5 Server at localhost failed to start.
    使用bootstrap中的bootstrapValidator,验证ckeditor富文本框不为空
    百度WebUploader上传图片,图片回显编辑,查看
    百度WebUploader上传图片
    做webapp静态页面的一些积累
    ztree插件的使用
    highcharts曲线图
    ajax的表单提交,与传送数据
    一条数据中需要遍历多条数据,页面遍历方法
    在页面中使用拼接字符串的方式显示动态加载的数据
  • 原文地址:https://www.cnblogs.com/lightwindy/p/8476983.html
Copyright © 2011-2022 走看看