[LeetCode] 81. Search in Rotated Sorted Array II
Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?Would this affect the run-time complexity? How and why?
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
).
Write a function to determine if a given target is in the array.
The array may contain duplicates.
Thought process:
The idea is similar to "154. Find Minimum in Rotated Sorted Array". If duplicates are allowed, worst case run-time will increase to O(n). Think about a case where the array contains a bunch of 1s and one 0, and we're searching for 0.
We eliminate the edge case where nums[left] == nums[right] at the beginning of the while loop. After that, the code is similar to the question where duplicates are not allowed. Pay attention to those <= and >= signs, because they do matter in some edge cases.
Solution:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 | public class Solution { public boolean search(int[] nums, int target) { if (nums.length == 0) { return false; } int left = 0; int right = nums.length - 1; int first = nums[0]; while (left + 1 < right) { if (nums[left] == nums[right]) { right--; continue; } int mid = left + (right - left) / 2; if (nums[mid] >= first) { if (nums[mid] < target || first > target) { left = mid; } else { right = mid; } } else { if (nums[mid] > target || first <= target) { right = mid; } else { left = mid; } } } if (nums[left] == target || nums[right] == target) { return true; } else { return false; } } } |
Time complexity:
Average case O(logn). Worst case O(n).
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