[LeetCode] Find First and Last Position of Element in Sorted

Problem

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]

Solution

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] res = {-1, -1};
        if (nums == null || nums.length == 0) return res;
        int start = 0, end = nums.length-1;
        while (nums[start] < nums[end]) {                   //don`t be equal
            int mid = start + (end-start)/2;
            if (nums[mid] < target) {
                start = mid+1;
            } else if (nums[mid] > target) {
                end = mid-1;
            } else {                                        //once nums[mid] == target:
                if (nums[start] != nums[mid]) start++;      //move start to lower bound (first position)
                else end--;                                 //move end to higher bound (last position)
            }
        }
        
        if (nums[start] == target && nums[end] == target) {
            res[0] = start;
            res[1] = end;
        }
        return res;
    }
}