34. Search for a Range

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].

二分查找的变形：

1、没有找到返回[-1, -1]

2、只存在1个返回[pos, pos]

3、存在多个，返回端点[leftPos, rightPos]

#include <iostream>#include <set>#include <map>#include <vector>#include <sstream>#include <string>#include <algorithm>#include <bitset>using namespace std;class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        vector<int>vec;        int leftPos = BinarySearchLeft(nums, target);        int rightPos = BinarySearchRight(nums, target);        if (nums[leftPos] != target)        {            vec.push_back(-1);            vec.push_back(-1);        }        else        {            vec.push_back(leftPos);            vec.push_back(rightPos);        }        return vec;    }private:int BinarySearchLeft(vector<int>& nums, int target)    {        int left = 0;        int right = nums.size() - 1;        while(left <= right)        {            int mid = left + ((right - left) >> 1);                      if (nums[mid] >= target)            {                right = mid - 1;            }            else            {                left = mid + 1;            }        }        return left;    }    int BinarySearchRight(vector<int>& nums, int target)    {        int left = 0;        int right = nums.size() - 1;        while(left <= right)        {            int mid = left + ((right - left) >> 1);                      if (nums[mid] > target)            {                right = mid - 1;            }            else            {                left = mid + 1;            }        }        return right;    }};int main(){Solution s;vector<int>vec{0,0,0,1,2,3};vector<int>v = s.searchRange(vec, 0);for (auto it = v.begin(); it != v.end(); it++){cout << *it << endl;}return 0;}

看，这个时间，估计得有更高效的算法了。