034 Search for a Range 搜索范围

给定一个已经升序排序的整形数组，找出给定目标值的开始位置和结束位置。
你的算法时间复杂度必须是 O(log n) 级别。
如果在数组中找不到目标，返回 [-1, -1]。
例如:
给出 [5, 7, 7, 8, 8, 10] 和目标值 8，
返回 [3, 4]。
详见：https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/

Java实现：

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int n=nums.length;
        int[] result = { -1, -1 };
        if(n==0||nums==null){
            return result;
        }
        int left = 0;
		int right = n - 1;
		while (left <= right) {
			int mid = (left + right)>>1;
			if (nums[mid] > target) {
				right = mid - 1;
			} else if (nums[mid] < target) {
				left = mid + 1;
			} else {
				int index = mid;
				while (index >= 0 && nums[index] == target) {
					--index;
				}
                result[0]=index+1;
				index = mid;
				while (index < nums.length && nums[index] == target) {
                    ++index;
				}
                result[1]=index-1;
				break;
			}
		}
        return result;
    }
}

C++实现：

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        //时间复杂度为O(logn)
		int n = nums.size();
		int l = 0, h = n-1, m;
		int start = -1, end = -1;
		while (l<=h)
        {
			m = l + ((h - l) >> 1);
			if (nums[m]>target)
            {
				h = m-1;
			}
			else if (nums[m]<target)
            {
				l = m + 1;
			}
			else
            {//找到以后，需要确定前后边界
				int index = m;
				while (index >= 0 && nums[index] == target)
                {
                    index--;
                }
				start = index + 1;
				index = m;
				while (index<n&&nums[index] == target)
                {
                    index++;
                }
				end = index - 1;
				break;
			}
		}
		return vector<int>{start,end};
    }
};