Algorithm-ticket purchase problem

I haven't posted a blog in August, let's practice dynamic planning.

The source of this topic is the NetEase written test

题目描述：现在有n个人排队买票，已知是早上八点开始卖票，这n个人买票有两种方式
第一种是每个人都可以单独去买自己的票，第i个人花费i秒。
第二种是每个人都可以选择和后面的人一起买票，第i个人和第i+1个人一共花费b[i]秒。
最后一个人只能和前一个一起买票或者自己单独购买，求花费的最短时间。

It can be seen that the problem of buying tickets is a typical dynamic programming problem. Why do you say that? Because the current minimum time depends on the previous ticket purchase decision.

After determining that it is a dynamic programming problem, our main task is to find the state transition equation.

The following example illustrates

        int[] a={
    
    3,5,7,6};
        int[] b={
    
    6,11,13};

When solving this problem, we are actually looking for the minimum value we accumulated in the array a from position 0 to position ptr

For example, when ptr=0, the current shortest time is obviously

dp[ptr]=a[0]

When ptr=1, the current shortest time is

dp[ptr]=min(a[0]+a[1],b[0])

Obviously, when ptr=2, the shortest time is min(dp[ptr-1]+a[ptr],dp[ptr-2]+b[ptr-1]).

dp[ptr]=min(dp[ptr-1]+a[ptr],dp[ptr-2]+b[ptr-1])

The rest can continue recursive

So we came to this conclusion

    /**
     * @param a 单独购买的时间
     * @param b 两个人一起买的时间
     * @return
     */
    public static int timeToClose(int[] a,int[] b){
    
    
        if (a.length==0||a.length==1){
    
    
            return a.length==0?0:a[0];
        }
        int[] dp=new int[a.length];
        dp[0]=a[0];
        dp[1]=Math.min(b[0],a[0]+a[1]);
        int ptr=2;
        while (ptr<a.length){
    
    
            //a[ptr]表示单独购买的花费时间，dp[ptr-1]表示从第0个人到第ptr-1个人总共花费的最小时间
            //b[ptr-1]表示第ptr个人和第ptr-1个人一起花费的时间，dp[ptr-2]表示从第0个人到第ptr-2个人总共花费的最小时间
            dp[ptr++]=Math.min(a[ptr]+dp[ptr-1],b[ptr-1]+dp[ptr-2]);
        }
        return dp[dp.length-1];
    }