LeetCode BinarySearch 34 Find First and Last Position of Element in Sorted Array

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 $O(logn)$.

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 1: two pointer

由于是sorted array，最直观的解法就是two pointer分别从两边向中间扫描，但是这样的解法复杂度是$O(n)$，不满足题目要求，如果想达到$O(logn)$，必须采用binary search。

solution 2: binary search

需要注意的是，target不存在的情况，以及target在nums中有duplicates的情况。 为了考虑duplicates，一发现nums[mid] == target，就把这个mid当作结果的candidates，分别向两边扫描，找到target第一次和最后一次出现的index

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

}

// public int BinarySearch(int[] nums, int target) {
//     int res = -1
//     if (nums == null || nums.length == 0) return res;
//     int lo = 0;
//     int hi = nums.length - 1;
    
//     while(lo <= hi) {
//         int mid = lo + (hi - lo) / 2;
//         if (nums[mid] > target) {
//             hi = mid - 1;
//         } else if (nums[mid] == target) {
//             hi = mid;
//         } else {
//             lo = mid + 1;
//         }
//     }
    
//     return nums[lo] == target ? lo:-1;
// }