[Leetcode]Search for a Range

Search for a Range My Submissions Question
Total Accepted: 64904 Total Submissions: 232560 Difficulty: Medium
Given a sorted array of integers, 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].

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一次找到第一个出现的位置，另一次找到最后一个出现的位置。进行替代即可

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