129 - 33 搜索旋转排序数组

题目

假设按照升序排序的数组在预先未知的某个点上进行了旋转。

( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。

搜索一个给定的目标值，如果数组中存在这个目标值，则返回它的索引，否则返回 -1 。

你可以假设数组中不存在重复的元素。

你的算法时间复杂度必须是 O(log n) 级别。

示例 1:

输入: nums = [4,5,6,7,0,1,2], target = 0 输出: 4

示例 2:

输入: nums = [4,5,6,7,0,1,2], target = 3 输出: -1

解答

怎么感觉可以直接用自带的api。。

或者排序一下两分搜索

var search = function(nums, target) {
  return nums.indexOf(target)
};

Runtime: 52 ms, faster than 87.17% of JavaScript online submissions for Search in Rotated Sorted Array.

Memory Usage: 33.9 MB, less than 30.77% of JavaScript online submissions for Search in Rotated Sorted Array.

自带api，效果还不错吼

看了题解，不用排序的，用两次两分法查找就行了

https://leetcode-cn.com/problems/search-in-rotated-sorted-array/solution/ji-jian-solution-by-lukelee/

就，找其中排序了的部分

• 如果nums[0] <= nums[mid]，说明0 - mid不包含旋转，有序。

  如果target还在这个区间里面，那么只需在里面查找

• 如果nums[mid] < nums[0]，说明0-mid包含旋转

  如果target小于mid，说明目标在旋转位置到mid之间

  如果target大于0，说明在0到旋转位置之间

var search = function(nums, target) {
  let left = 0,
    right = nums.length - 1
  while (left < right) {
    let mid = (left + right) >>> 1
    if ((nums[0] > target) ^ (nums[0] > nums[mid]) ^ (target > nums[mid])) {
      left = mid + 1
    } else {
      right = mid
    }
  }
  return nums[left] === target ? left : -1
};

Runtime: 60 ms, faster than 44.56% of JavaScript online submissions for Search in Rotated Sorted Array.

Memory Usage: 34.1 MB, less than 7.69% of JavaScript online submissions for Search in Rotated Sorted Array.

不过while出来的left和right始终是保持相等的。。不用做题解中的判断

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        if not nums:
            return -1
        left = 0
        right = len(nums)-1
        while left < right:
            mid = (left + right) >> 1
            if (nums[0] > target) ^ (nums[0] > nums[mid]) ^ (target > nums[mid]):
                left = mid+1
            else:
                right = mid
        return left if nums[left] == target else -1

Runtime: 48 ms, faster than 79.30% of Python3 online submissions for Search in Rotated Sorted Array.

Memory Usage: 13.9 MB, less than 6.29% of Python3 online submissions for Search in Rotated Sorted Array.

go居然不能用异或判断bool。。。所以只能用!=来代替。。

func search(nums []int, target int) int {
    if len(nums) == 0 {
        return -1
    }
    left := 0
    right := len(nums) - 1
    for left < right {
        mid := (left + right) >> 1
        if ((nums[0] > target) != (nums[0] > nums[mid])) != (target > nums[mid]) {
            left = mid + 1
        } else {
            right = mid
        }
    }
    if nums[left] == target {
        return left
    } else {
        return -1
    }
}

Runtime: 0 ms, faster than 100.00% of Go online submissions for Search in Rotated Sorted Array.

Memory Usage: 2.6 MB, less than 50.00% of Go online submissions for Search in Rotated Sorted Array.