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].

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Have you met this question in a real interview?

思路 ： 肯定是 binary search .最后再向两边扩展。

public class Solution {    public int[] searchRange(int[] A, int target) {        int[] ret = new int[2];        ret[0] = -1;        ret[1] = -1;        if(A.length < 1)            return ret;        int left = 0;        int right = A.length - 1;        boolean isFind = false;        while(left <= right){            int mid = (left + right) / 2;            if(A[mid] == target){                left = mid;                right = mid;                while(left >= 0 && A[left] == target)                    left--;                while(right < A.length && A[right] == target)                    right++;                ret[0] = left + 1;                ret[1] = right - 1;                return ret;            }else if(A[mid] < target)                left = mid + 1;            else                right =mid - 1;        }        return ret;    }}