search/search-in-rotated-sorted-array-ii
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2023-12-01
Search in Rotated Sorted Array II
描述
Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
Write a function to determine if a given target is in the array.
分析
允许重复元素，则上一题中如果A[left] <= A[mid],那么[left,mid]为递增序列的假设就不能成立了，比如[1,3,1,1,1]。
既然A[left] <= A[mid]不能确定递增，那就把它拆分成两个条件：
- 若
A[left] < A[mid]，则区间[left,mid]一定递增 - 若
A[left] == A[mid]确定不了，那就left++，往下看一步即可。
代码
// Search in Rotated Sorted Array II
// Time Complexity: O(n)，Space Complexity: O(1)
public class Solution {
public boolean search(int[] nums, int target) {
int first = 0, last = nums.length;
while (first != last) {
final int mid = first + (last - first) / 2;
if (nums[mid] == target)
return true;
if (nums[first] < nums[mid]) {
if (nums[first] <= target && target < nums[mid])
last = mid;
else
first = mid + 1;
} else if (nums[first] > nums[mid]) {
if (nums[mid] < target && target <= nums[last-1])
first = mid + 1;
else
last = mid;
} else
//skip duplicate one
first++;
}
return false;
}
};
// Search in Rotated Sorted Array II
// Time Complexity: O(n)，Space Complexity: O(1)
class Solution {
public:
bool search(const vector<int>& nums, int target) {
int first = 0, last = nums.size();
while (first != last) {
const int mid = first + (last - first) / 2;
if (nums[mid] == target)
return true;
if (nums[first] < nums[mid]) {
if (nums[first] <= target && target < nums[mid])
last = mid;
else
first = mid + 1;
} else if (nums[first] > nums[mid]) {
if (nums[mid] < target && target <= nums[last-1])
first = mid + 1;
else
last = mid;
} else
//skip duplicate one
first++;
}
return false;
}
};