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问题描述

我需要找到数组中数字之间的最长路径(从大到小).我尝试编写 recursive 函数并得到 java.lang.StackOverflowError,但由于缺乏知识,我不明白为什么会发生这种情况.

I need to find the longest way (from bigger to lower) between numbers in array.I've tried to write recursive function and got java.lang.StackOverflowError, but for the lack of knowledge I didn't understand why this happened.

首先,我已经初始化数组并用随机数填充它:

Firstly, I've initialize array and fill it with random numbers:

public long[] singleMap = new long[20];
for (int i = 0; i < 20; i++) {
        singleMap[i] = (short) random.nextInt(30);
    }

然后,我尝试找到最长的倒数路线(例如 { 1, 4, 6, 20, 19, 16, 10, 6, 4, 7, 6, 1 ...} ) 并返回这些数字的计数.

Then, I try to find the longest route of counting down numbers (e.g. { 1, 4, 6, 20, 19, 16, 10, 6, 4, 7, 6, 1 ...} ) and to return the count such numbers.

 public int find(long[] route, int start) {
    if (route[start] > route[start + 1]) {
        find(route, start++);
    } else {
        return start;
    }
    return start;
}

所以这是日志:

 08-23 13:06:40.399 4627-4627/itea.com.testnotification I/dalvikvm:   threadid=1: stack overflow on call to    Litea/com/testnotification/MainActivity;.find:ILI
 08-23 13:06:40.399 4627-4627/itea.com.testnotification I/dalvikvm:   method requires 36+20+12=68 bytes, fp is 0x4189e318 (24 left)
 08-23 13:06:40.399 4627-4627/itea.com.testnotification I/dalvikvm:   expanding stack end (0x4189e300 to 0x4189e000)
 08-23 13:06:40.400 4627-4627/itea.com.testnotification I/dalvikvm: Shrank stack (to 0x4189e300, curFrame is 0x418a3e88)
 08-23 13:06:40.400 4627-4627/itea.com.testnotification D/AndroidRuntime: Shutting down VM
 08-23 13:06:40.400 4627-4627/itea.com.testnotification W/dalvikvm: threadid=1: thread exiting with uncaught exception (group=0x41a8ed40)
 08-23 13:06:40.414 4627-4627/itea.com.testnotification E/AndroidRuntime: FATAL EXCEPTION: main
                                                                         Process: itea.com.testnotification, PID: 4627
                                                                     java.lang.StackOverflowError
                                                                         at itea.com.testnotification.MainActivity.find(MainActivity.java:46)
                                                                         at itea.com.testnotification.MainActivity.find(MainActivity.java:46)

我感谢任何解释,因为所有相关问题都没有帮助我.如果我的功能有问题,请纠正或解释.

I appreciate any explanation because all related issues didn't help me. If there is a problem in my function, please correct or explain.

编辑

我忘了说,我使用 for 来检查每个点"的最长路径

I forgot to say, that I use for to check the longest way from each "point"

  for (int i = 0; i < singleMap.length - 1; i++) {
        int x = find(singleMap, i);
        System.out.println("steps = " + x);
    }

推荐答案

您需要保持当前的最大查找到现在,以及当前值.所以改成如下:

You need to mantain the current max find up to now, and the current value.So change it as follow:

public int find(int[] route, int start, int max, int currentMax) {
    if (currentMax > max) {
        max = currentMax;
    }
    if (start == route.length - 1) {
        return max;
    }
    if (route[start] > route[start + 1]) {
        return find(route, start + 1, max, currentMax + 1);
    }
    return find(route, start + 1, max, 1);
}

并以开头调用它

find(route, 0, 1, 0);

第二种选择是不递归地重写它:


A second alternative is to rewrite it without recursion:

public int find(int[] route) {
    int max = 1;
    int currentMax = 1;
    for (int i = 0; i < route.length - 1; i++) {
        if (route[i] > route[i + 1]) {
            currentMax++;    // If next element is lower increment currentMax
            if (currentMax > max) {
                max = currentMax;   // If currentMax is the new max update max
            }

        } else {
            currentMax = 1;   // If next element is not lower restart from 1
        }
    }
    return max;
}

并称之为

find(route);

这篇关于递归java上的堆栈溢出错误的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

08-13 12:49