• Leetcode-34-Search for a Range-(Medium)


    这道题借助二分查找算法来查找目标值的index

    然后向前和向后分别搜索起始边界

    注意开始排除异常值优化速度

    #!/usr/local/bin/python3
    # -*- coding: utf-8 -*-
    __author__ = 'author'
    
    
    class Solution(object):
        def searchRange(self, nums, target):
            """
            :type nums: List[int]
            :type target: int
            :rtype: List[int]
            """
            length = len(nums)
            if len(nums) == 0 or nums[0] > target or nums[length - 1] < target:
                return [-1, -1]
            index = self.binary_search(nums, 0, length - 1, target)
            if index == -1:
                return [-1, -1]
            else:
                start = index
                end = index
                for i in range(index - 1, -1, -1):
                    if nums[i] == target:
                        start = i
                    else:
                        break
                for i in range(index + 1, length):
                    if nums[i] == target:
                        end = i
                    else:
                        break
                return [start, end]
    
        #二分查找算法
        def binary_search(self, nums, start, end, targrt):
            if start > end:
                return -1
            mid = start + (end - start)//2
            if nums[mid] > targrt:
                return self.binary_search(nums, start, mid - 1, targrt)
            elif nums[mid] < targrt:
                return self.binary_search(nums, mid + 1, end, targrt)
            else:
                return mid
    

    另外的一种思路是寻找 target-1 和 target+1的所在位置的索引,这两个值可能不存在,那么需要相应的修改二分查找算法

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  • 原文地址:https://www.cnblogs.com/doudouyoutang/p/6289495.html
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