leetcode34 - Find First and Last Position of Element in Sorted Array - medium

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]

 

Constraints:

• 0 <= nums.length <= 10^5
• -10^9 <= nums[i] <= 10^9
• nums is a non decreasing array.
• -10^9 <= target <= 10^9

用sheep的模板，找start pos的话是模板2-r总是返回true，找end pos是模板1-l总是返回true。

找到start pos后要check一下此时nums[r]是否是target，不是的话就是找不到。

开始找end pos时把r更新回array尾端，l其实不用改，end pos至少也和此时的l相等。

 

实现：O(logn)

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        
        vector<int> res{-1, -1};
        if (nums.empty())
            return res;
        
        int l = 0, r = nums.size()-1;
        while (l < r){
            int m = l + (long long)r >> 1;
            if (nums[m] >= target){
                r = m;
            }else{
                l = m+1;
            }
        }
        
        if (nums[r] != target)
            return res;
        
        res[0] = r;
        
        r = nums.size()-1;
        while (l < r){
            int m = l + (long long)r + 1 >> 1;
            if (nums[m] <= target){
                l = m;
            }else{
                r = m-1;
            }
        }
        
        res[1] = l;
        
        return res;
        
    }
};