leetcode打印零偶奇问题显示错误Time Limit Exceeded答案

【问题标题】：Print Zero Even Odd problem of leetcode shows error Time Limit Exceededleetcode打印零偶奇问题显示错误Time Limit Exceeded
【发布时间】：2020-03-15 18:41:50
【问题描述】：

我为打印零偶奇问题的leetcode问题尝试了以下代码。在测试最简单的输入时，它向我显示 Time Limit Exceeded 错误。有人可以提出一些建议吗？ https://leetcode.com/problems/print-zero-even-odd

class ZeroEvenOdd {

    private int n;

    public ZeroEvenOdd(int n) {
        this.n = n;
    }

    // printNumber.accept(x) outputs "x", where x is an integer.
    boolean z = true;
    public void zero(IntConsumer printNumber) throws InterruptedException {
        for(int i = 1; i <= n; i++) {
            if(z) {
                printNumber.accept(0);
            }
            z = false;
            odd = !odd;
            try { notifyAll(); } catch(Exception e){}
            do { try { wait(); } catch(Exception e){ } } while(!z);
        }
    }
    boolean odd = false;
    public void even(IntConsumer printNumber) throws InterruptedException {
        for(int i = 1; i <= n; i++) {
            do { try { wait(); } catch(Exception e){ } } while(z || odd);
            if(!odd && !z && i%2 == 0) {
                printNumber.accept(i);
            }
            z = true;
            try { notifyAll(); } catch(Exception e){}
        }
    }

    public void odd(IntConsumer printNumber) throws InterruptedException {
        for(int i = 1; i <= n; i++) {
            do { try { wait(); } catch(Exception e){ } } while(z || !odd);
            if(odd && !z && i%2 == 1) {
                printNumber.accept(i);
            }
            z = true;
            try { notifyAll(); } catch(Exception e){}
        }
    }
   }

【问题讨论】：

标签： java multithreading

【解决方案1】：

尝试使用synchronized：代码看起来更简单

class ZeroEvenOdd {
    private int n;

    public ZeroEvenOdd(int n) {
        this.n = n;
    }

    private enum STATUS{
        ZERO, ODD, EVEN;
    }

    private STATUS status = STATUS.ZERO;
    private int x = 0;
    private Object lock = new Object();

    // printNumber.accept(x) outputs "x", where x is an integer.
    public void zero(IntConsumer printNumber) throws InterruptedException {
        while(x<=n){
            switch(status){
                case ZERO:
                    synchronized(lock){
                        if(x < n){
                            printNumber.accept(0);    
                        }
                        if(x%2==0){
                            status = STATUS.ODD;
                        }else{
                            status = STATUS.EVEN;
                        }
                        x++;
                    }
                    break;
            }
        }
    }

    public void even(IntConsumer printNumber) throws InterruptedException {
        while(x<=n){
            switch(status){
                case EVEN:
                    if (x<=n){
                        printNumber.accept(x);
                    }
                    status = STATUS.ZERO;
                break;
            }
        }
    }

    public void odd(IntConsumer printNumber) throws InterruptedException {

        while(x<=n){
            switch(status){
                case ODD:
                    synchronized(lock){
                        if(x <= n){
                            printNumber.accept(x);
                        }
                        status = STATUS.ZERO;
                    }
                    break;
            }
        }
    }
}

如需synchronized的完整参考，请访问this

【讨论】：