81 Search in Rotated Sorted Array II
1. Question
Follow up for "Search in Rotated Sorted Array": What ifduplicatesare 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 become4 5 6 7 0 1 2
).
Write a function to determine if a given target is in the array.
The array may contain duplicates.
2. Implementation
(1) Binary Search
思路: 这题和33题的区别在于数组存在重复,但基本思路还是一样,由于二分查找的运用前提是数组要有序的,所以分成4种情况:
(1) nums[mid] 等于 target,找到target,直接返回true
(2) nums[mid] < nums[end], [mid, end]区间是有序的
如果nums[mid] < target <= nums[end], 在[mid + 1, end]区间继续做二分搜索
否则在[start, mid - 1]做二分搜索
(3) nums[mid] > nums[end], 由于数组时rotate一次,所以[start, mid]是有序的
如果nums[start <= target < nums[mid], 在[start, mid - 1]区间做二分搜索
否则在[mid + 1, end]二分搜索
(4) nums[mid]等于nums[end],此时无法判断方向,end往左移一步
class Solution {
public boolean search(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return false;
}
int start = 0, end = nums.length - 1, mid = 0;
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (nums[mid] == target) {
return true;
}
else if (nums[mid] < nums[end]) {
if (nums[mid] < target && target <= nums[end]) {
start = mid + 1;
}
else {
end = mid - 1;
}
}
else if (nums[mid] > nums[end]) {
if (nums[start] <= target && target < nums[mid]) {
end = mid - 1;
}
else {
start = mid + 1;
}
}
else {
--end;
}
}
return nums[start] == target || nums[end] == target;
}
}
3. Time & Space Complexity
时间复杂度O(logn),空间复杂度O(1)
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