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  • 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;
        }
    };
    

      

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