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There is an integer array nums sorted in ascending order (with distinct values).
Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].
Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
You must write an algorithm with O(log n) runtime complexity.
Example 1:
<pre>Input: nums = [4,5,6,7,0,1,2], target = 0 Output: 4 </pre>
Example 2:
<pre>Input: nums = [4,5,6,7,0,1,2], target = 3 Output: -1 </pre>
Example 3:
<pre>Input: nums = [1], target = 0 Output: -1 </pre>
Constraints:
1 <= nums.length <= 5000-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup>- All values of
numsare unique. numsis an ascending array that is possibly rotated.-10<sup>4</sup> <= target <= 10<sup>4</sup>
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题目大意：
有序数组旋转了几位，求target的下标
算法思路：
N/A
注意事项：
- 四种情况： 左半有序且target在这个有序区间内，左边有序的所有其他情况，右半有序且target在这个有序区间，右半有序的所有其他情况
Python代码：
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17def search(self, nums: List[int], target: int) -> int:
start, end = 0, len(nums) - 1
while start + 1 < end:
mid = start + (end - start) // 2
if nums[start] < nums[mid] and nums[start] <= target <= nums[mid]: # remember
end = mid
elif nums[start] < nums[mid]:
start = mid
elif nums[mid] < nums[end] and nums[mid] <= target <= nums[end]:
start = mid
else:
end = mid
if nums[start] == target:
return start
if nums[end] == target:
return end
return -1
Java代码：
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29public int search(int[] nums, int target) {
if(nums == null || nums.length == 0)
return -1;
int start = 0, end = nums.length - 1;
while(start + 1 < end) {
int mid = start + (end - start) / 2;
if(nums[mid] > nums[start]) { // the left segment is increasing
// pay attention to equal
if(nums[start] <= target && target <= nums[mid]) //the [start, mid] is increasing
end = mid;
else
start = mid;
}
else {
if(nums[mid] <= target && target <= nums[end]) // the [mid, end] is increasing
start = mid;
else
end = mid;
}
}
if(nums[start] == target)
return start;
if(nums[end] == target)
return end;
return -1;
}
算法分析：
时间复杂度为O(logn)，空间复杂度O(1)
