LeetCode 34. Search for a Range

 Second solution is my favourite actually. But it seems that some interviewer prefer to do it recursively.

#include <iostream>#include <vector>using namespace std;/*    Given a sorted arry of integers. Find the starting and ending position    of a given target value.    algorithm's run time should be in order of O(lgn)    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];*/// First solution. O(lgn) but worst time O(n).vector<int> searchRange(vector<int>& nums, int target) {    int left = 0, right = nums.size() - 1;    vector<int> res;    while(left <= right) {        int mid = (left + right) / 2;        if(nums[mid] == target) {            int left = mid;            int right = mid;            while(nums[left] == target) {left--;}            while(nums[right] == target) {right++;}            if((nums[left+1] == target) && (nums[right-1] == target)) {                res.push_back(left+1);                res.push_back(right-1);                return res;            }        } else if(nums[mid] < target) left = mid + 1;        else right = mid - 1;    }    return {-1, -1};}

int searchRangeRight(vector<int>& nums, int target) {    int left = 0, right = nums.size() - 1;    while(left < right) {        int mid = (left + right + 1) / 2; // here should plus 1, make it moving.        if(nums[mid] > target) right = mid - 1;        else left = mid;    }    if(nums[left] == target) return left;    return -1;}int searchRangeLeft(vector<int>& nums, int target) {    int left = 0, right = nums.size() - 1;    while(left < right) {        int mid = (left + right) / 2;        if(nums[mid] < target) left = mid + 1;        else right = mid;    }    if(nums[right] == target) return right;    return -1;}

// Second solution. always O(lgn) even with repeations.vector<int> searchRangeII(vector<int>& nums, int target) {    int left = 0, right = nums.size() - 1;    int leftIndex = searchRangeLeft(nums, target);    int rightIndex = searchRangeRight(nums, target);    return {leftIndex, rightIndex};}int searchRangeIII(vector<int>& nums, int left, int right, int target, bool flag) {    if(left > right) return -1;    int mid = (left + right) / 2;    if(nums[mid] == target) {        int pos = flag ? searchRangeIII(nums, left, mid - 1, target, flag) : searchRangeIII(nums, mid + 1, right, target, flag);        return pos == -1 ? mid : pos;    } else if(nums[mid] < target)        searchRangeIII(nums, mid + 1, right, target, flag);    else        searchRangeIII(nums, left, mid - 1, target, flag);}

// Let's do it recursively....vector<int> searchRangeIII(vector<int>& nums, int target) {    int left = 0, right = nums.size() - 1;    bool leftFlag = true;    int leftIndex = searchRangeIII(nums, left, right, target, leftFlag);    int rightIndex = searchRangeIII(nums, left, right, target, !leftFlag);    return {leftIndex, rightIndex};}int main(void) {    vector<int> nums{5, 7, 7, 8, 8, 10};    vector<int> res = searchRangeIII(nums, 7);    for(int i = 0; i < res.size(); ++i) {        cout << res[i] << endl;    }}