34. Search for a Range

1.Question

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].

2.Code

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        int last_label = nums.size() - 1;        int left = 0, right = last_label;        int mid;        vector<int> res(2, -1);        if(last_label == 0)//考虑数组长度为1的情况        {            if(nums[0] == target) {res[0] = 0; res[1] = 0;}            return res;        }        while(left <= right)//先找到一个target        {            mid = left + 0.5*(right - left);            if(nums[mid] == target) break;            else if(nums[mid] > target) right = mid - 1;            else left = mid + 1;        }        if(left > right) return res;        int left1 = left, right1 = mid;        int left2 = mid, right2 = right;        int flag = 1;        while(flag)//找左边边缘的target        {            mid = left1 + 0.5*(right1 - left1);            if(nums[mid] == target)            {                if(mid == 0 || nums[mid-1] != target) {res[0] = mid; flag = 0;}                else right1 = mid - 1;            }            else left1 = mid + 1;        }        flag = 1;        while(flag) //找右边边缘的target        {            mid = left2 + 0.5*(right2 - left2);            if(nums[mid] == target)            {                if(mid == last_label || nums[mid+1] != target) {res[1] = mid; flag = 0;}                else left2 = mid + 1;            }            else right2 = mid - 1;        }        return res;    }};

3.Note

a. 这个题还是binary search的思想，只是需要分别找两个target（左右两边各一个）。把细节处理好就行。