LeetCode #34 - Find First and Last Position of Element in Sorted Array

タイトルの説明:
Given an array of integers  nums  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] .
Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

O(log n)の時間的複雑度で秩序配列でなければならないので,二分探索を考慮する.ターゲット要素が見つかったら、それぞれ両側に検索し、その要素の最初の位置と最後の位置を見つけます.

class Solution {
public:
    vector searchRange(vector& nums, int target) {
        int begin=0;
        int end=nums.size()-1;
        int mid=(begin+end)/2;
        while(begin<=end)
        {
            if(nums[mid]==target)
            {
                int i=mid;
                int j=mid;
                while(nums[i-1]==target&&i>0) i--;
                while(nums[j+1]==target&&j result;
                result.push_back(i);
                result.push_back(j);
                return result;
            }
            else if(target>nums[mid])
            {
                begin=mid+1;
                mid=(begin+end)/2;
            }
            else if(target(2,-1);
    }
};