剑指offer编程题解法汇总37-数字在排序数组中出现的次数

题目描述

统计一个数字在升序数组中出现的次数。

示例1

输入

[1,2,3,3,3,3,4,5],3

返回值

4

想法自然是二分查找，代码如下：

/**
     * [],3
     * 4
     */
    public int GetNumberOfK(int[] array, int k) throws Exception {
        try{
            int binarysearchIndex = binarysearch(array, k);
            int num = 1;
            int middle = binarysearchIndex;
            while ((--middle >= 0) && array[middle] == k) {
                num++;
            }
            while ((++binarysearchIndex <= (array.length - 1)) && array[binarysearchIndex] == k) {
                num++;
            }
            return num;
        }catch (Exception e){
        }

        return 0;
    }

    /**
     * 二分查找找位置
     */
    public int binarysearch(int[] array, int k) throws Exception {

        if (k < array[0] || k > array[array.length - 1]) {
            throw new Exception("error");
        }
        int start = 0;
        int end = array.length - 1;
        int middle = 0;
        while (true) {
            middle = (start + end) / 2;
            if (k == array[middle]) {
                break;
            }
            if (start == end) {
                throw new Exception("error");
            }
            if ((end - start) == 1) {
//                直接判断
                if (array[start] == k) {
                    return start;
                } else if (array[end] == k) {
                    return end;
                } else {
                    throw new Exception("error");
                }
            }
            if (array[middle] > k) {
                end = middle;
            } else {
                start = middle;
            }
        }
        return middle;
    }

测试代码：

  public static void main(String[] args) {
        Solution solution = new Solution();
        int[] arrays = {1, 1, 1, 1, 1,2};
        int binarysearch = 99;
        try {
            binarysearch = solution.GetNumberOfK(arrays, 3);
        } catch (Exception e) {
            e.printStackTrace();
        }
        System.out.println(binarysearch);
    }