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
class Solution {
public int search(int[] nums, int target) {
}
}
说，假设有一个升序排序的数组，可能在某个点旋转。
找到目标值就返回下标，找不到返回-1。
可以假设没有重复的数字。
时间复杂度在O(log n)以内。
要在O(log n)以内首先想到进行折半查找，但是进行旋转后，就不是全部有序的了，这点要注意。
以下为代码：
public int search(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return -1;
}
int n = nums.length;
int left = 0;
int right = n - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
return mid;
}
if (nums[mid] > target) {
if (nums[mid] >= nums[0]) {
if (target >= nums[0]) {
right = mid - 1;
} else {
left = mid + 1;
}
} else {
right = mid - 1;
}
} else {
if (nums[mid] >= nums[0]) {
left = mid + 1;
} else {
if (target <= nums[n - 1]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
}
}
return -1;
}
