递归排序方法不会在基本情况下停止

时间:2018-11-27 23:10:38

标签: java

对于学校,我正在尝试编写一个递归排序算法。但是由于某种原因,它并没有停止在基本情况下,将不胜感激。错误是

  

java.lang.StackOverflowError

这是方法:

 public static List recursiveMergeSort(List<Integer> list1){
    List<Integer> list = list1;
    List<Integer> listR = new ArrayList<>();
   List<Integer> listL = new ArrayList<>();


     int counter =0;

    if(list.size() == 1){
        return list;
    } else {

        for (int i = 0; i <list.size()/2-1; i++){
            listL.add(list.get(i));
        }

        for(int i = list.size()/2; i  < list.size()-1; i++) {
            listR.add(list.get(i));
        }

        recursiveMergeSort(listL);
        recursiveMergeSort(listR);

        list = merge(listL, listR);
        counter++;
    }
    return (list);
}

public static List merge(List<Integer> listL, List<Integer> listR){
    int leftIndex = 0;
    int rightIndex = 0;
    int listIndex = 0;
    List<Integer> list = new ArrayList<>();
    while(leftIndex < listL.size() && rightIndex < listR.size()){
        if(listL.get(leftIndex) <= listR.get(leftIndex)){
            listR.add(listIndex, listL.get(leftIndex));
            leftIndex = leftIndex + 1;

        }else {
            list.add(list.get(listIndex), listR.get(rightIndex));
            rightIndex = rightIndex +1;
        }
        listIndex = listIndex + 1;
    }
    if(leftIndex < listL.size()){
        copyRest(listL, leftIndex, list, listIndex);
    } else {
        copyRest(listR, rightIndex, list, listIndex);
    }
    return (list);
}

感谢您的帮助!

1 个答案:

答案 0 :(得分:1)

问题出在两个内部的for循环中:

    for (int i = 0; i <list.size()/2-1; i++){ // should be i<list.size()/2
        listL.add(list.get(i));
    }


    for(int i = list.size()/2; i  < list.size()-1; i++) { //should be i  < list.size()
        listR.add(list.get(i));
    }

最初,当您提供大小为5的输入时,listR获得两个元素(具有索引2和3的元素),并且在对listR的递归调用中,不会执行任何for循环,因此递归调用将一直进行直到你得到一个错误