34. Search for a Range(C++)

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].

Solution:

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> m_vec(2,-1);
        bool s_flag=false,e_flag=false;
        for(int i=0,j=nums.size()-1;i<=j;) {
            if(nums[i]==target) {
                m_vec[0]=i;
                s_flag=true;
            }else i++;
            if(nums[j]==target){
                m_vec[1]=j;
                e_flag=true;
            }else j--;
            if(s_flag&&e_flag) break;
        }
        return m_vec;
    }
};

  

原文地址:https://www.cnblogs.com/devin-guwz/p/6696456.html