题目：统计一个数字在排序数组中出现的次数。

思路一：简单暴力的解法一。。。

package offer;

public class Solution999 {
	public int GetNumberOfK(int[] array, int k) {
		int count = 0;
		for (int i = 0; i < array.length; i++) {
			if (array[i] == k)
				count++;
		}
		return count;
	}
}

思路二：二分查找法

package offer;

public class Solution999 {

	public int GetNumberOfK(int[] array, int k) {
		int length = array.length;
		if (length == 0) {
			return 0;
		}
		int firstK = getFirstK(array, k, 0, length - 1);
		int lastK = getLastK(array, k, 0, length - 1);
		if (firstK != -1 && lastK != -1) {
			return lastK - firstK + 1;
		}
		return 0;
	}

	// 递归写法
	private int getFirstK(int[] array, int k, int start, int end) {
		if (start > end) {
			return -1;
		}
		int mid = (start + end) >> 1;
		if (array[mid] > k) {
			return getFirstK(array, k, start, mid - 1);
		} else if (array[mid] < k) {
			return getFirstK(array, k, mid + 1, end);
		} else if (mid - 1 >= 0 && array[mid - 1] == k) {
			return getFirstK(array, k, start, mid - 1);
		} else {
			return mid;
		}
	}

	// 循环写法
	private int getLastK(int[] array, int k, int start, int end) {
		int length = array.length;
		int mid = (start + end) >> 1;
		while (start <= end) {
			if (array[mid] > k) {
				end = mid - 1;
			} else if (array[mid] < k) {
				start = mid + 1;
			} else if (mid + 1 < length && array[mid + 1] == k) {
				start = mid + 1;
			} else {
				return mid;
			}
			mid = (start + end) >> 1;
		}
		return -1;
	}
}