Binary Search - 33. Search in Rotated Sorted Array

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发布于 2020-09-23 17:41:34

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33. Search in Rotated Sorted Array

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

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

思路：

题目意思是找出一个翻转过的数组中目标元素的下标，使用binary search来做，根据最左边和中间的元素来区分中间元素的左边是翻转过的还是没有翻转过的。

代码：

go：

func search(nums []int, target int) int {

    if nums == nil || len(nums) == 0 {
        return -1
    }
    
    left := 0
    right := len(nums)-1
    
    for left <= right {
        mid := left + (right - left) / 2
        if nums[mid] == target {
            return mid
        } else if nums[mid] < nums[left]  {
            // 左边是rotated sorted array 
            if target > nums[mid] && target <= nums[right] {
                left = mid + 1
            } else {
                right = mid - 1
            }
        } else {
            // 右边是rotated sorted array 左边是sorted array
            if target < nums[mid] && target >= nums[left] {
                right = mid - 1
            } else {
                left = mid + 1;
            }
        }
    }
    
    return -1
}

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原始发表：2019年10月10日，如有侵权请联系 cloudcommunity@tencent.com 删除

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