数组练习题10--在排序数组中查找数字I

题意：
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

方法1：哈希表
思路：统计出现次数，直接返回次数即可

class Solution {
    public int search(int[] nums, int target) {
    	Map<Integer,Integer> record=HashMap<>();
    	for(int i=0;i<nums.length;i++){
    		record.put(nums[i],record.getOrDefault(nums[i],0)+1);
    	}
    	return record.getOrDefault(target,0);
    }
}

方法2：二分
思路：找到target的左右端点，将端点相减+1得到出现次数：

class Solution {
    public int search(int[] nums, int target) {
        int left=binaySearchLeftBound(nums,target);
        int right=binarySearchRightBound(nums,target);
        if(left!=-1)
            return right-left+1;
        return 0;
    }
    public int binaySearchLeftBound(int[] nums,int target){
        int low=0,high=nums.length-1;
        while(low<=high){
            int mid=low+(high-low)/2;
            if (nums[mid]==target){
                if(mid==0||nums[mid-1]<target)
                    return mid;
                high=mid-1;
            }
            else if(nums[mid]>target){
                high=mid-1;
            }
            else if(nums[mid]<target){
                low=mid+1;
            }
        }
        return -1;
    }
    public int binarySearchRightBound(int[] nums,int target){
        int low=0,high=nums.length-1;
        while(low<=high){
            int mid=low+(high-low)/2;
            if (nums[mid]==target){
                if(mid==nums.length-1||nums[mid+1]>target)
                    return mid;
                low=mid+1;
            }
            else if(nums[mid]>target){
                high=mid-1;
            }
            else if(nums[mid]<target){
                low=mid+1;
            }
        }
        return -1;
    }
}