LeetCode

题目：

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：

二分查找

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);
    }

}