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

posted @ 2017-01-16 14:26  兜兜有糖的博客  阅读(176)  评论(0编辑  收藏  举报