[北京|结业弟子]CSS-曾添|2017年08月18日的日报-技能树.IT修真院

发表于： 2017-08-18 20:18:55

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今天，前端测试出现一个bug，postman和浏览器测试接口正常，页面测试json格式有一堆斜杠，经过仔细的debug发现了这个bug的原因。

出现bug的原因是两个：一个是前端页面发送的请求和postman发送请求的区别是，前端发送请求中多了"Accept: application/json"这一项，第二个原因是添加了Jackson依赖后，会自动向Spring中增加一个jsonHttpMessageConverter。这两个原因共同导致了这个bug。

具体来说，在使用了@ResponseBody之后，Spring判断使用哪一种HttpMessageConverter的机制是这样的：

首先遍历请求头中 Accept: 后面所有的类型，所以遍历的第一个类型就是application/json。

其次遍历所有的HttpMessageConverter，只要遍历出符合application/json的HttpMessageConverter，直接使用此HttpMessageConverter，不管你Response中具体的格式是什么。

把Jackson这个依赖注解掉后，问题解决，当然应该有自定义HttpMessageConverter的方法去解决这个问题。

做了一些简单的算法练习题。正确答案都看不懂。

1.找到一个数组中出现了奇数次的那个唯一的数。

package august18;
import java.util.HashMap;
public class FindOdd {
    public static void main(String[] args) {
        int [] arr = {1,2,1};
        System.out.println(findItBest(arr));
    }
    public static int findIt(int[] A) {
        HashMap<Integer,Integer> map = new HashMap<>();
        for (int i : A){
            if (!map.containsKey(i)){
                map.put(i, 1);
            }
            else {
                map.put(i, map.get(i) + 1);
            }
        }
        for (int i : map.keySet()){
            if (map.get(i) % 2 != 0){
                return i;
            }
        }
        return 0;
    }
    public static int findItBest(int[] A) {
        int xor = 0;
        for (int i = 0; i < A.length; i++) {
            xor = xor ^ A[i];
            //xor ^= A[i];
        }
        return xor;
    }
}

2.一个数字问题，附两个标准答案。

persistenceBest2(39) == 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
// and 4 has only one digit

persistenceBest2(999) == 4 // because 9*9*9 = 729, 7*2*9 = 126,
// 1*2*6 = 12, and finally 1*2 = 2

persistenceBest2(4) == 0 // because 4 is already a one-digit number

package august18;
/**
* Created by ZengTian on 2017/8/18.
* Write a function, persistenceBest2, that takes in a positive parameter num and returns its multiplicative persistenceBest2, which is the number of times you must multiply the digits in num until you reach a single digit.
For example:
persistenceBest2(39) == 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
// and 4 has only one digit
persistenceBest2(999) == 4 // because 9*9*9 = 729, 7*2*9 = 126,
// 1*2*6 = 12, and finally 1*2 = 2
persistenceBest2(4) == 0 // because 4 is already a one-digit number
*/
public class PersistentBugger {
    public static void main(String[] args) {
        System.out.println(persistence(999L));
    }
    private static int count = 0;
    private static int persistence(long n){
        count = 0;
        calculation(n);
        return count;
    }
    private static int calculation(long n) {
        if (n < 10){
            String s = String.valueOf(n);
            return Integer.parseInt(s);
        }
        count ++;
        return calculation(middle(n));
    }
    private static long middle(long n){
        if (n < 1){
            return 1;
        }
        long left = n / 10;
        long right = n - (n / 10) * 10;
        return middle(left) * right;
    }
    //标准答案
    public static int persistenceBest(long n) {
        long m = 1, r = n;
        if (r / 10 == 0)
            return 0;
        for (r = n; r != 0; r /= 10)
            m *= r % 10;
        return persistenceBest(m) + 1;
    }
    public static int persistenceBest2(long n) {
        int times = 0;
        while (n >= 10) {
            n = Long.toString(n).chars().reduce(1, (r, i) -> r * (i - '0'));
            times++;
        }
        return times;
    }
}

3.排队问题，前面数组是队伍的人需要的时间，后面数字代表有几个柜台，计算一共多少时间。这个和标准答案一样。

package august18;
import java.util.Collections;
import java.util.PriorityQueue;
/**
* Created by ZengTian on 2017/8/18.
* There is a queue for the self-checkout tills at the supermarket. Your task is write a function to calculate the total time required for all the customers to check out!
The function has two input variables:
customers: an array (list in python) of positive integers representing the queue. Each integer represents a customer, and its value is the amount of time they require to check out.
n: a positive integer, the number of checkout tills.
The function should return an integer, the total time required.
EDIT: A lot of people have been confused in the comments. To try to prevent any more confusion:
There is only ONE queue, and
The order of the queue NEVER changes, and
Assume that the front person in the queue (i.e. the first element in the array/list) proceeds to a till as soon as it becomes free.
The diagram on the wiki page I linked to at the bottom of the description may be useful.
So, for example:
queueTime([5,3,4], 1)
// should return 12
// because when n=1, the total time is just the sum of the times
queueTime([10,2,3,3], 2)
// should return 10
// because here n=2 and the 2nd, 3rd, and 4th people in the
// queue finish before the 1st person has finished.
queueTime([2,3,10], 2)
// should return 12
N.B. You should assume that all the test input will be valid, as specified above.
P.S. The situation in this kata can be likened to the more-computer-science-related idea of a thread pool, with relation to running multiple processes at the same time: https://en.wikipedia.org/wiki/Thread_pool
*/
public class TheSupermarketQueue {
    public static void main(String[] args) {
        int [] arr = {2,3,10};
        System.out.println(solveSuperMarketQueue(arr, 2));
    }
    public static int solveSuperMarketQueue(int[] customers, int n) {
        PriorityQueue<Integer> queue = new PriorityQueue<>();
        for (int i = 0; i < n; i++){
            queue.add(0);
        }
        for (int i : customers){
            int p = queue.poll();
            queue.add(p + i);
        }
        return Collections.max(queue);
    }
}