[LeetCode]034. Search for a Range

Given a sorted array of integers, 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].

Solution: use binary search, when find the target, do left shift and right shift find the range.

Running Time: O(lg(n)), worst case: O(n).

public class Solution {
    public int[] searchRange(int[] A, int target) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int[] result = {-1,-1};
        int len = A.length;
        int left = 0;
        int right = len - 1;
        while(left<=right){
            int mid = (left+right)/2;
            if(A[mid] == target){
                int min = mid;
                int max = mid;
                while(min-1>=0 && A[min] == A[min-1]){
                    min--;
                }
                while(max+1<len && A[max] == A[max+1]){
                    max++;
                }
                result[0] = min;
                result[1] = max;
                return result;
            }else if(A[mid] < target){
                left = mid + 1;
            }else{
                right = mid - 1;
            }
        }
        return result;
    }
}

2nd solution:

A solution of o(logn). Do two runs of binary search for the index of largest element strictly less than target, and the index of largest element strictly less than target+1. The range is between this two indices. 

To be continued