剑指offer--数字在排序数组中出现的次数（java）

1. 遍历一遍数组，进行查找，这个方法没有利用排序数组这个条件。时间复杂度较高
   时间复杂度：O(n)

       //一层循环遍历数组，省略
   

   2.利用排序数组这个条件，通过二分法来找到，找到第一个k的索引和最后一个数组的索引。
   时间复杂度：O(logn)

       public int GetNumberOfK(int [] array , int k) {
       	if(array == null || array.length ==0)
       		return 0;
       	int firstIndex = getKIndex(array, k, 0, array.length-1, 1);
       	int lastIndex = getKIndex(array, k, 0, array.length-1, 0);
       	if(firstIndex==-1 || lastIndex==-1)
       		return 0;
       	return lastIndex-firstIndex+1;
       }
   
   	private int getKIndex(int[] array, int k, int low, int high, int isFirst) {
   		int middleIndex = (low+high)/2;
   		int middleValue = array[middleIndex];
   		int newLow = low;
   		int newHigh = high;
   		if(low>high) {
   			return -1;
   		}
   		if(k==middleValue) {
   			if(isFirst == 0) {
   			if(middleIndex == array.length-1 || array[middleIndex+1] != k)
   				return middleIndex;
   			newLow = middleIndex +1;
   			}
   			else {
   				if(middleIndex == 0 || array[middleIndex-1] != k)
   					return middleIndex;
   				newHigh = middleIndex -1;
   			}
   		}
   		else if(k<middleValue) {
   			newHigh = middleIndex - 1;
   		}
   		else {
   			newLow = middleIndex + 1;
   		}
   		return getKIndex(array, k, newLow, newHigh, isFirst);
   	
   	}