Leetcode 34. Find First and Last Position of Element in Sorted Array

Description: Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. If target is not found in the array, return [-1, -1].

Follow up: Could you write an algorithm with O(log n) runtime complexity?

Link: 34. Find First and Last Position of Element in Sorted Array

Examples:

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]

Example 3:
Input: nums = [], target = 0
Output: [-1,-1]

思路: 首先用正常的二分查找流程找到target所在的任意index，如何找不到就返回[-1,-1]，如果找到了，当前mid = index所在的区间[l,r]一定包含目标值的start，end，再在这里区间中分别确定start 和 end 的位置。也是利用二分查找。start可能在[l,mid-1]里，end可能在[mid+1,r]里，过一个例子就知道start=sl,end=er.

class Solution(object):
    def searchRange(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        if not nums: return [-1,-1]
        n = len(nums)
        l, r = 0, n-1
        while l <= r:
            mid = int((l+r)/2)
            if nums[mid] == target:
                sl = l
                sr = mid-1
                while sl <= sr:
                    sm = int((sl+sr)/2)
                    if nums[sm] == target:
                        sr = sm - 1
                    else:
                        sl = sm + 1
                el = mid+1
                er = r
                while el <= er:
                    em = int((el+er)/2)
                    if nums[em] == target:
                        el = em + 1
                    else:
                        er = em - 1
                return [sl, er] 
            elif target > nums[mid]:
                l = mid + 1
            else:
                r = mid - 1
        return [-1,-1]

日期: 2021-04-10 所有人，所有事都在告诉我，没有人能随随便便成功，要坚持啊！