剑指 Offer 53 - I. 在排序数组中查找数字 I

转载自Leet Code《剑指Offer》

题目描述

统计一个数字在排序数组中出现的次数。

示例 1：

输入：nums = [5,7,7,8,8,10], target = 8 输出：2

示例 2：

输入：nums = [5,7,7,8,8,10], target = 6 输出：0

提示：

• 0 <= nums.length <= 105
• -109 <= nums[i] <= 109
• nums是一个非递减数组
• -109 <= target <= 109

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我的代码

{.line-numbers}

class MySolutionOffer53I {
    public int search(int[] nums, int target) {
        
    	int count = 0;
        if (nums==null||nums.length==0) return count;
    	
    	int index0 = BinarySearch(nums, 0, nums.length, target);
    	if (index0>=nums.length||nums[index0]!=target) return count;
    	
    	int index1 = BinarySearch(nums,index0,nums.length,target+1);
    	return index1-index0;
    }
    
	// array[first, last); array[index]>=value
	static int BinarySearch(int []array, int first, int last, int value)
	{
		while (first<last)
		{
			int mid = first+(last-first)/2;
			if (array[mid]<value)
				first = mid+1;
			else
				last=mid;
		}
		return first; //or last.
	}
}