[LeetCode] (medium) 34. Find First and Last Position of Element in Sorted Array

https://leetcode.com/problems/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]

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 二分搜索上下界，用的邓俊辉的理解方法，很直观。

搜索上界时，hi及hi的右侧满足>target，lo的左侧满足<=target，因此当lo==hi时，lo-1即为target应在位置的上界

搜索下界时，lo及lo的左侧满足<target，hi的右侧满足>=target，因此当lo==hi时，hi+1即为target应在位置的下界

注意在搜索下界时要确保每一次迭代问题的潜在解空间会缩小，即每一步都必须要移动两侧边界，所以mid = (lo+hi+1)/2

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int hi = search_up(nums, target);
        if(hi == -1 || nums[hi] != target) return {-1, -1};
        int lo = search_down(nums, target);
        return {lo, hi};
    }
    
    int search_up(vector<int>& nums, int target){
        int lo = 0;
        int hi = nums.size();
        int mid;
        while(lo < hi){
            mid = (lo+hi)/2;
            if(nums[mid] > target) hi = mid;
            else lo = mid+1;
        }
        return lo-1;
    }
    
    int search_down(vector<int>& nums, int target){
        int lo = -1;
        int hi = nums.size()-1;
        int mid;
        while(lo < hi){
            mid = (lo+hi+1)/2;
            if(nums[mid] < target) lo = mid;
            else hi = mid-1;
        }
        return lo+1;
    }
};