Leetcode: 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].

思路：两次二分搜索，分别找start pos和end pos。

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