leetcode : Search for a Range

Posted 2020-08-31 notesbuddy

Given an array of integers 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].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

 

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        
        int[] result = new int[2];
		result[0] = result[1] = -1;
		
		if(nums == null || nums.length == 0){
			return result;
		}
		
		int start = 0;
		int end = nums.length - 1;
		int mid;
		while(start < end - 1){
			mid = start + (end - start) / 2;
			
			if(nums[mid] == target){
			    end = mid;
			}else if(nums[mid] > target){
			    end = mid;
			}else{
			    start = mid;
			}
		}
		
		if(nums[start] == target){
		    result[0] = start;
		}else if(nums[end] == target){
		    result[0] = end;
		}else{
		    return result;
		}
		
			
		start = 0;
		end = nums.length - 1;
		while(start < end - 1){
			 mid = start + (end - start) / 2;
			
			if(nums[mid] == target){
			    start = mid;
			}else if(nums[mid] > target){
			    end = mid;
			}else{
			    start = mid;
			}
		}
		
		if(nums[end] == target){
		    result[1] = end;
		}else if(nums[start] == target){
		    result[1] = start;
		}else{
		    return result;
		}
		
		return result;
    }
}