Search for a Range(在升序数组中找到目标值的左右边界)leetcode34

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最新推荐文章于 2025-05-20 18:28:23 发布

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这道题要求在升序排列的整数数组中，寻找给定值的起始和结束位置。算法的时间复杂度需为O(log n)。当目标值不在数组中时，返回空数组。

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Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

 public int[] searchRange(int[] nums, int target) {
        int [] result=new int[]{-1,-1};
        if(nums.length<=0)
            return result;
        int index=searchIndex(nums,0,nums.length-1,target);
        System.out.println(index);
        if(index==-1)
            return result;
        int leftIndex=index;
        while(leftIndex!=0){
        if(leftIndex>0&&nums[leftIndex-1]<nums[leftIndex])
            break;
        leftIndex=searchIndex(nums,0,leftIndex-1,target);
        }
        int rightIndex=index;
        while(rightIndex!=nums.length-1){
        if(nums[rightIndex+1]>nums[rightIndex])
            break;
        rightIndex=searchIndex(nums,rightIndex+1,nums.length-1,target);
        }
        result[0]=leftIndex;
        result[1]=rightIndex;
        return result;
    }
    int searchIndex(int[] nums,int low,int high,int target){
        while(low<=high){
        int mid=low+(high-low)/2;
        if(nums[mid]==target)
            return mid;
        else if(nums[mid]<target)
            low=mid+1;
        else
            high=mid-1;
        }
        return -1;
    }