34. Search for a Range

题目

Given an array of integers sorted in ascending order, 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].

思路

这种查找类的题目，而且原数组还有序，基本的思路就是二分查找，找到相等的值，再从两边扩散查找

代码

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int l=0,r=nums.size()-1,mid;
        while(l<=r)
        {
            mid = (l+r)/2;
            if(nums[mid]>target)
                r = mid-1;
            else if(nums[mid]<target)
                l=mid+1;
            else 
                break;
        }
        if(l<=r)
        {
            int temp = mid;
            while(mid>=0)
            {
                if(nums[mid]==target)
                    mid--;
                else
                    break;
            }
            while(temp<=nums.size()-1)
            {
                if(nums[temp]==target)
                    temp++;
                else
                    break;
            }
            return {mid+1,temp-1};
        }
        else
            return {-1,-1};
    }
};