🔗:代码随想录:二分查找的算法讲解:有关left<=right和left<right的区别

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int n=nums.length;
        int l=0,h=n-1;
        if(nums==null){
            return null;    
        }
        if(n==0){
           return new int[]{-1,-1};    
        }
        if(target<nums[0]||target>nums[n-1]){
            return new int[]{-1,-1}; 
        }
        int m=(l+h)/2;
        boolean Find=false;
        while(l<=h){
            m=(l+h)/2;
            if(nums[m]<target){
                l=m+1;
            }
            else if(nums[m]>target){
                h=m-1;
            }else{
                Find=true;
                break;
            }
        }
        if(nums[m]==target&&Find){
            int i=m,j=m;
                while(i>0&&nums[i]==target&&nums[i-1]==nums[i]){
                    i--;
                }
                while(j+1<n&&nums[j]==target&&nums[j+1]==nums[j]){
                    j++;
                }
                return new int[]{i,j};
        }
        return new int[]{-1,-1};    
    }
    
}