zoukankan      html  css  js  c++  java
  • search in rotated sorted array

    descripte problem:

      there is an array,at the first it is sorted,but for som reason, it was rotated at some pivot unknown to you beforehand.then you should find the target num in the array,and return the index

    i.e:

    [4,5,6,1,2,3]

    [1]

    return 

    4

    solution:

      you should treat the array as two aescend array,which named part1,part2.Then you can use the binary search to find the low ,high and middle.

      first, calculate the middle = (low +high)/2

      second,if nums[middle] == target, then middle is the index which we find ;else judge which part is the middle belonged to, part1 or part2

      third, if the middle belong to part1, you should know that whether nums[low] <= target <= nums[middle], if ture it turns out that target must exits between nums[low] and nums[middle],so you just need make high = middele -1;if it is false, the target is impossible to exits between nums[low] and nums[middle], so you just need make low = middle +1.

      fourth, if the middle belong to part2, you have to find if nums[middle] <= target <= nums[high], if ture it turns  out that target must exits between nums[middle] and nums[high], so you just need make low = middle +1

    ;if false, you should make high = middle - 1

    there is my python code

    class Solution(object):
        def search(self, nums, target):
            """
            :type nums: List[int]
            :type target: int
            :rtype: int
            """
            low = 0
            high = len(nums)-1
            while low<=high:
                middle = (low+high)/2
                if nums[middle]==target:
                    return middle
                if nums[middle] >= nums[low]:
                    if target < nums[middle] and target >= nums[low]:
                        high = middle - 1
                    else:
                        low = middle + 1
                else:
                    if target <= nums[high] and target >= nums[middle]:
                        low = middle + 1
                    else:
                        high = middle - 1
                
            return -1
                

    of course, you can just use the order search, but it has be proved that the exceed time is not stable

    there is the simple solution

    class Solution(object):
        def search(self, nums, target):
            """
            :type nums: List[int]
            :type target: int
            :rtype: int
            """
            for i in range(len(nums)):
                if nums[i] == target:
                    return i
            return -1
  • 相关阅读:
    H5图片裁剪升级版(手机版)
    仿IOS 开关按钮
    JS 数字转换为大写金额
    Unity UGUI——遮罩效果(Mask)
    AdTime:多屏时代下传统媒体的鼓起
    C语言中随机数相关问题
    在vc中使用xtremetoolkit界面库-----安装及环境配置
    初识HTML 5:关于它的三个三
    移动中间件产品的解决方式
    Android ListView 常见问题与使用总结
  • 原文地址:https://www.cnblogs.com/whatyouknow123/p/6755388.html
Copyright © 2011-2022 走看看