计算已排序数组中数字的出现次数

Question

我的老师给了我下一个任务:

On a sorted array, find the number of occurrences of a number.
The complexity of the algorithm must be as small as possible.

这就是我的想法:

public static int count(int[] a, int x)
{
    int low = 0, high = a.length - 1;

    while( low <= high )
    {
        int middle = low + (high - low) / 2;

        if( a[middle] > x ) {
            // Continue searching the lower part of the array
            high = middle - 1;
        } else if( a[middle] < x ) {
            // Continue searching the upper part of the array
            low = middle + 1;
        } else {
            // We've found the array index of the value
            return x + SearchLeft(arr, x, middle) + SearchRight(arr, x, middle);
        }
    }

    return 0;
}

SearchLeft并SearchRight迭代数组,直到数字不显示.

我不确定我是否已经为这个问题编写了更快的代码,我希望看到其他意见.

编辑:经过评论和答案的一些帮助,这是我目前的尝试:

public static int count(int[] array, int value)
{
    return SearchRightBound(array, value) - SearchLeftBound(array, value);
}

public static int SearchLeftBound(int[] array, int value)
{
    int low = 0, high = array.length - 1;

    while( low < high )
    {
        int middle = low + (high - low) / 2;

        if(array[middle] < value) {
            low = middle + 1;
        }
        else {
            high = middle;
        }
    }

    return low;
}

public static int SearchRightBound(int[] array, int value)
{
    int low = 0, high = array.length - 1;

    while( low < high )
    {
        int middle = low + (high - low) / 2;

        if(array[middle] > value) {
            high = middle;
        }
        else {
            low = middle + 1;
        }
    }

    return low;
}

Answer 1

SearchLeft和SearchRight迭代数组,直到数字不显示.

这意味着如果整个数组都填充了目标值,那么算法就是O(n).

O(log n)如果你二元搜索第一次和最后一次出现,你可以做到最坏的情况x.

// search first occurrence
int low = 0, high = a.length - 1;
while(low < high) {
    int middle = low + (high-low)/2;
    if (a[middle] < x) {
        // the first occurrence must come after index middle, if any
        low = middle+1;
    } else if (a[middle] > x) {
        // the first occurrence must come before index middle if at all
        high = middle-1;
    } else {
        // found an occurrence, it may be the first or not
        high = middle;
    }
}
if (high < low || a[low] != x) {
    // that means no occurrence
    return 0;
}
// remember first occurrence
int first = low;
// search last occurrence, must be between low and a.length-1 inclusive
high = a.length - 1;
// now, we always have a[low] == x and high is the index of the last occurrence or later
while(low < high) {
    // bias middle towards high now
    int middle = low + (high+1-low)/2;
    if (a[middle] > x) {
        // the last occurrence must come before index middle
        high = middle-1;
    } else {
        // last known occurrence
        low = middle;
    }
}
// high is now index of last occurrence
return (high - first + 1);