算法---巧妙的二分查找

統計一個數字在排序數組中出現的次數。

思路一：傳統解法，採用二分查找第一個K和最後一個K；

public class GetNumberOfK {

	public static void main(String[] args) {
		
		int array[] = {1,2,3,3,3,3,3};
		int k = 3;
		int sum = getNumberOfK(array,k);
		System.out.println(sum);
	}

	public static int getNumberOfK(int[] array, int k) {
	if(array.length<=0)
			return 0;
		
		int lastk = biSearchlast(array,k,0,array.length-1);
		int firstk = biSearchFirst(array,k,0,array.length-1);
		if(lastk != -1 && firstk != -1)
			return  lastk - firstk + 1;
		return 0;
	}
    
	public static int biSearchlast(int[] array, int k, int start, int end) {
		int mid = 0;
		int length = array.length - 1;
		while(start <= end)
		{
			mid = (start+end)>>1;
			if(array[mid] < k)
			{
				start = mid + 1;
			}
			else if(array[mid] > k)
			{
				end = mid - 1;
			}
			else if(mid+1<=length && array[mid+1]==k){
				start = mid + 1;
			}else {
				return mid;
			}
		}
		return -1;
	}

	public static int biSearchFirst(int[] array, int k, int start, int end) {
		int mid = 0;
		while(start <= end){
			mid = (end + start)>>1;
			if(array[mid] > k)
			{
				end = mid - 1;
			}
			else if(array[mid] < k)
			{
				start = mid + 1;
			}
			else if(mid - 1 >= 0 && array[mid-1] == k)
			{
				end = mid - 1;
			}
			else {
				return mid;
			}
		}
		return -1;
	}
	}

思路二：看到有序，想到二分查找。
因爲data中都是整數，所以可以稍微變一下，不是搜索k的兩個位置，而是搜索k-0.5和k+0.5
這兩個數應該插入的位置，然後相減即可。

public class GetNumberOfK {

	public static void main(String[] args) {
		
		int array[] = {1,2,3,3,3,3,3};
		int k = 3;
		int sum = getNumberOfK(array,k);
		System.out.println(sum);
	}

	public static int getNumberOfK(int[] array, int k) {
		if(array.length<=0)
			return 0;
		
		return biSearch(array,k+0.5) - biSearch(array,k-0.5);
	}

	public static int biSearch(int[] array, double d) {
		int min = 0;
		int max = array.length-1;
		int mid = 0;
		while(min <= max)
		{
			mid = (max+min)/2;
			if(array[mid] < d)
			{
				min = mid + 1;
			}
			else if(array[mid] > d)
			{
				max = mid - 1;
			}
		}
		return max;
	}
}