Find First and Last Position of Element in Sorted Array

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

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

You must write an algorithm with O(log n) runtime complexity.

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]

Example 3:

Input: nums = [], target = 0
Output: [-1,-1]

Constraints:

• 0 <= nums.length <= 105
• -109 <= nums[i] <= 109
• nums is a non-decreasing array.
• -109 <= target <= 109

思路：两个method，模板走起。

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