search/search-in-rotated-sorted-array
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2023-12-01
Search in Rotated Sorted Array
描述
Suppose a sorted array 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.
分析
一个有序数组被循环右移，只可能有一下两种情况：
7 │
6 │
─────┼───────────
│ 5
│ 4
│ 3
│ 2
│ 1
7 │
6 │
5 │
4 │
3 │
───────────┼───────────
│ 2
│ 1
本题依旧可以用二分查找，难度主要在于左右边界的确定。仔细观察上面两幅图，我们可以得出如下结论：
如果A[left] <= A[mid],那么[left,mid] 一定为单调递增序列。
代码
// Search in Rotated Sorted Array
// Time Complexity: O(log n)，Space Complexity: O(1)
public class Solution {
public int 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 mid;
if (nums[first] <= nums[mid]) {
if (nums[first] <= target && target < nums[mid])
last = mid;
else
first = mid + 1;
} else {
if (nums[mid] < target && target <= nums[last-1])
first = mid + 1;
else
last = mid;
}
}
return -1;
}
};
// Search in Rotated Sorted Array
// Time Complexity: O(log n)，Space Complexity: O(1)
class Solution {
public:
int 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 mid;
if (nums[first] <= nums[mid]) {
if (nums[first] <= target && target < nums[mid])
last = mid;
else
first = mid + 1;
} else {
if (nums[mid] < target && target <= nums[last-1])
first = mid + 1;
else
last = mid;
}
}
return -1;
}
};