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};
    }
};