leetcode [34] Find First and Last Position of Element in Sorted Array

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]

 

题目大意：

在给定的排序数组中寻找目标数出现的重复区间，如果没有这个数，则返回{-1，,1}

 

解法：

根据题目的时间复杂度采用二分查找进行搜索。

C++

class Solution {
public:
    void searchRangeCore(vector<int>nums,int start,int end,int target,vector<int>&res){
        if(end<start) return;
        if(nums[start]>target||nums[end]<target) return;
        int mid=(end-start)/2+start;
        if(nums[mid]==target){
            res[0]=(res[0]==-1?mid:min(res[0],mid));
            res[1]=max(res[1],mid);
        }
        searchRangeCore(nums,start,mid-1,target,res);
        searchRangeCore(nums,mid+1,end,target,res);
    }

    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int>res={-1,-1};
        searchRangeCore(nums,0,nums.size()-1,target,res);

        return res;
    }
};