二分查找--34. 在排序数组中查找元素的第一个和最后一个位置

你的算法时间复杂度必须是 O(log n) 级别。

如果数组中不存在目标值，返回 [-1, -1]。

示例 1:

输入: nums = [5,7,7,8,8,10], target = 8
输出: [3,4]

代码实现

class Solution {
public:
    int left_index(vector<int>& nums,int target)
    {
        int begin = 0;
        int end = nums.size() - 1;
        while(begin <= end)
        {
           int mid = (begin + end) / 2;
            if(nums[mid] == target)
            {
                if(mid == 0 || nums[mid - 1] < target)
                {
                    return mid;
                }
                end = mid - 1;
            }
            else if(target < nums[mid])
            {
                end = mid - 1;
            }
            else if(target > nums[mid])
            {
                begin = mid + 1;
            }
        }
        return -1;
        }
        int right_index(vector<int>& nums,int target)
        {
            int begin = 0;
            int end = nums.size() - 1;
            while(begin <= end)
            {
              int mid = (begin + end) / 2;
                if(nums[mid] == target)
                {
                    if(mid == nums.size() - 1 || nums[mid + 1] > target)
                    {
                        return mid;
                    }
                    begin = mid + 1;
                }
                else if(target < nums[mid])
                {
                    end = mid - 1;
                }
                else if(target > nums[mid])
                {
                    begin = mid + 1;
                }
            }
            return -1;
        }    
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> result;
        result.push_back(left_index(nums,target));
        result.push_back(right_index(nums,target));
        return result;
    }
};