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
// nums = [4,5,6,7,0,1,2], target = 0 [a1,a2,a3,..,am,b1,b2,..,bn] b1<b2<..<bn<a1<a2<..<an //1. target < a1 2. target > a1 3. target > bn && target < a1 func search(nums []int, target int) int { if len(nums) ==0{ return -1 } left, right := 0, len(nums)-1 a1, bn := nums[0], nums[len(nums)-1] // 1.没有旋转的情况 if a1<bn{ return simpleSearch(nums,0,target) } //2. b1<b2<..<bn<a1<a2<..<an if target>bn&&target<a1{ return -1 } for left <= right { if nums[left] == target { return left } if nums[right] == target { return right } mid := (left + right) / 2 switch { case nums[mid] == target: return mid case nums[mid] <target: if nums[mid]>a1{ left = mid +1 a1 = left }else{ if target<a1{ return simpleSearch(nums[mid:right],mid,target) }else{ right = mid -1 bn =right } } case nums[mid]>target: if nums[mid]<a1{ right = mid -1 bn = right }else{ if target>bn{ return simpleSearch(nums[left:mid],left,target) }else{ left = mid +1 } } } } return -1 } func simpleSearch(nums []int,offset, target int) int{ left,right := 0, len(nums)-1 for left < right { mid:=(left+right)/2 if nums[left]==target{ return offset+left } if nums[right]==target{ return offset+right } switch { case nums[mid]==target : return offset+ mid case nums[mid] < target : left = mid + 1 case nums[mid] > target: right = mid -1 } } return -1 }