Question

我想找到给定数字x在排序数组中的首次出现。这是我到目前为止的方法：

public static int firstOccur(int[] A, int x, int start, int end){

    int first = start;
    int last = end;
    int result = -1;

    while (first <= last){
        int mid = (first+last)/2;

        if(A[mid] == x){
            result = mid;   
            firstOccur(A, x, first, mid-1);
            return result;
        }
        else if(A[mid] < x){
            first = mid+1;
            result = firstOccur(A, x, first, last);
        }
        else if(A[mid] > x){
            last = mid-1;
            return firstOccur(A, x, first, last);
        }
    }
    return result;
}

public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);
    System.out.print("Please enter the value you are looking for: ");
    int val = sc.nextInt();

    int[] arr = {1,2,2,2,5};
    System.out.println("The first occurence of " + val + " is at index " + firstOccur(arr, val, 0, arr.length-1));

}

运行代码时，该函数适用于数字1 & 5 and anything added。不幸的是，在提交x=2时，它返回索引2，这是不正确的。我错过了一个小细节吗？

Answer 1

您不在这里考虑递归函数的返回值

if(A[mid] == x){
    result = mid;   
    firstOccur(A, x, first, mid-1);
    return result;
}

将其更改为

if(A[mid] == x) {
    result = mid;   
    int maybeResultToLeft = firstOccur(A, x, first, mid-1);
    if (maybeResultToLeft == -1) {
       return result;
    }
    return maybeResultToLeft;
}

或者单线

return maybeResultToLeft == -1? result : maybeResultToLeft;

我们需要选择当前（x）元素（如果存在-x）左侧的元素（maybeResultToLeft != -1）

递归查找数字在SORTED数组中的首次出现

1 个答案: