34. Find First and Last Position of Element in Sorted Array

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]

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] res = {-1,-1};
        int left = 0, right = nums.length - 1;
        int l = -1, r = -1;
        while(left <= right){
            int mid = left + (right-left)/2;
            if(nums[mid] == target){
                l = mid;
                r = mid;
                while((l-1) > -1 && nums[l-1] == target){
                    l--;
                }
                while((r+1) < nums.length && nums[r+1] == target){
                    r++;
                }
                break;//！！
            }
            else{
                if(nums[mid] > target){
                    right = mid - 1;
                }
                else left = mid + 1;
            }
        }
        res[0] = l;
        res[1] = r;
        return res;
    }
}

Binary Search。之所以要break是因为一旦所有符号条件的数选完就不会进入到下面的else，也不会更新right和left，所以会有time limit error。因此当得到l和r就应该break。

总结：Using two pointers again, first is to use binary search to find the first target, then make l = mid and r = mid, expand in two directions and break. Then l is the first index, r is the last index. Including the situations that nums.length == 0/1 or only 1 target in array.