LeetCode, leetcodeoj

LeetCode, leetcodeoj

Question:

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 orderO(LogN).

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].

Ideas:

Binary Search

package array;public class SearchForARange {    public int[] searchRange(int[] nums, int target) {        int[] res = { -1, -1 };        int n;        if (nums == null || (n = nums.length) == 0) return res;                int index = -1;        int start = 0;        int end = n - 1;        while (start <= end) {            int mid = (end - start) / 2 + start;            if (nums[mid] == target) {                index = mid;                break;            } else if (nums[mid] < target) {                start = mid + 1;            } else {                end = mid - 1;            }        }                   res[0] = index;           res[1] = index;           while (res[0] > 0 && nums[res[0] - 1] == target) --res[0];           while (res[1] < n - 1 && nums[res[1] + 1] == target) ++res[1];                return res;    }        public static void main(String[] args) {        // TODO Auto-generated method stub        int[] nums = { 5, 7, 7, 8, 8, 10 };        SearchForARange s = new SearchForARange();        for(int i : s.searchRange(nums, 10))        System.out.println(i);    }}