# Leetcode [No.34] Find First and Last Position of Element in Sorted Array (Medium)

## 題目

Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.

If target is not found in the array, return [-1, -1].

You must write an algorithm with O(log n) runtime complexity.

## 思路
+ 一看到這個題目，有修過資料結構的同學應該可以馬上知道至少需要使用binarySearch。
+ 但binarySearch有非常多的眉角，我在這裡使用左閉右開的定義!
+ 先找到其中之一的結果後，再去重新計算最左的值，因為一array有可能會是重複的值。
+ 計算完左邊換計算右邊
+ 最後return結果就行了～

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int front = 0; int end = nums.size();
        vector<int>sol;
        int ans = (binarySearch(nums,front,end,target));
        if(ans == -1) {sol.emplace_back(-1);sol.emplace_back(-1);return sol;}
        int new_front = 0; int new_end = 0;
        for(int i = front; i <= ans; i++)
        {
            if(nums[i] == target)
            {
                new_front = i;
                break;
            }
        }
        
        for (int i = end-1; i >= ans; i--)
        {
            if(nums[i] == target)
            {
                new_end = i;
                break;
            }
        }
        
        sol.emplace_back(new_front);sol.emplace_back(new_end);
        return sol;
    }
    
    int binarySearch(vector<int>& nums, int front, int end, int target)
    {
        while(front < end)
        {
            int mid = (front+end)/2;
            if(nums[mid] == target)
            {
                return mid;
            }
            else if (nums[mid] > target)
            {
                end = mid;
            }
            else // nums[mid] < target
            {
                front = mid+1;
            }
        }
        return -1;

    }
};

### 解法分析
+ time complexity: O(lgn+n)

### 執行結果