Search for a Range.java

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].

public class Solution {
    public int[] searchRange(int[] A, int target) {
        int low = findLow(A, target, 0, A.length - 1);
        int high = findHigh(A, target, 0, A.length - 1);
        int[] ret = new int[2];
        ret[0] = low;
        ret[1] = high;
        return ret;
    }
    
    private int findLow(int[] A, int target, int l, int r) {
        int mid = 0;
        int ret = -1;
        while (l <= r) {
            mid = (l + r) / 2;
            if (A[mid] == target) {
                ret = mid;
                int next = findLow(A, target, l, mid - 1);
                if (next != -1) {
                    ret = next;
                }
                break;
            } else if (A[mid] < target) {
                l = mid + 1;
            } else {
                r = mid - 1;
            }
            
        }
        return ret;
    }

    private int findHigh(int[] A, int target, int l, int r) {
        int mid = 0;
        int ret = -1;
        while (l <= r) {
            mid = (l + r) / 2;
            if (A[mid] == target) {
                ret = mid;
                int next = findHigh(A, target, mid + 1, r);
                if (next != -1) {
                    ret = next;
                }
                break;
            } else if (A[mid] < target) {
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        return ret;
    }
}