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SearchInRotatedSortedArray.java
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SearchInRotatedSortedArray.java
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// Link: https://leetcode.com/problems/search-in-rotated-sorted-array/
// Binary search idea
// 0ms, 100%
class Solution {
public int search(int[] nums, int target) {
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
return mid;
}
if (nums[mid] >= nums[left]) {
if (target < nums[mid] && target >= nums[left]) {
right = mid - 1;
} else {
left = mid + 1;
}
} else {
if (target > nums[mid] && target <= nums[right]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
}
return -1;
}
}
/*
* Time complexity: O(logn)
* Due to the property of the binary search, each iteration, we rule out the half of the array
*
* Space complexity: O(1)
*
* Notes:
* Using the binary search idea, each time we rule out the half.
* Take [4,5,6,7,0,1,2] as an example:
* - When mid is at 4~7, if 4 <= target < mid, we can rule out the right half of mid; otherwise
* the left half can be ruled out.
* - When mid is at 0~2, if mid < target <= 2, we can rule out the left half of mid; otherwise
* the right half can be ruled out.
*/