动态规划实战1-leetcode 983.Minimum Cost For Tickets

前言：最简单的动态规划比如爬楼梯、斐波那契数列就不做举例分析了。我们来看一下场景相对复杂的题目。

确定状态 dp[i]表示到第i天所用的最少的cost

状态转移方程（这道题不难写）

dp[i]=dp[i-1] 如果在第i天没有travel

dp[i]= min {dp[i-1]+costs[0],dp[i-7]+costs[1],dp[i-30]+costs[2]}

初始条件和边界条件

Arrays.fiil(dp,Integer.MAX_VALUE)

dp[0]=0;

我们来看下这个表达式：dp[i-30] 如果i<30说明此时我们应该从第0天就买为期30天的票，而不是dp[负数]。

状态转移方程可以修改表示为如下：

dp[i]=dp[i-1] 如果在第i天没有travel

dp[i]= min {dp[i-1]+costs[0],dp[max{i-7,0}]+costs[1],dp[max{i-30,0}]+costs[2]}

代码如下：

class Solution {public int mincostTickets(int[] days, int[] costs) {if(days==null||days.length==0)return 0;int maxday=days[days.length-1];int []dp = new int [maxday+1];int []req = new int [maxday+1];for(int i=0;i<days.length;i++){req[days[i]]=1;}Arrays.fill(dp,Integer.MAX_VALUE);dp[0]=0;for(int i=1;i<=maxday;i++){if(req[i]==0){dp[i]=dp[i-1];}else{dp[i]=Math.min(dp[i],dp[i-1]+costs[0]);dp[i]=Math.min(dp[i],dp[Math.max(i-7,0)]+costs[1]);dp[i]=Math.min(dp[i],dp[Math.max(i-30,0)]+costs[2]);}}return dp[maxday];}
}

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