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
40
41
42
43
44
45
46
47
48
|
// 81. Search in Rotated Sorted Array II
#include "leetcode.h"
/*
* there is an integer array 'nums' sorted in non-decreasing order (not
* necessarily with distinct values). before being passed to your function,
* 'nums' is rotated at an unknown pivot index 'k(0 <= k < nums.length)' such
* that the resulting array is '[nums[k], nums[k + 1], ..., nums[n - 1],
* nums[0], nums[1], ..., nums[k - 1]]' (0-indexed). for example
* '[0,1,2,4,4,4,5,6,6,7]' might be rotated at a pivot index 5 and become
* '[4,5,6,6,7,0,1,2,4,4]'. given the array 'nums' after the rotation and an
* integer 'target', return true if 'target' is in 'nums' of 'false' if it is
* not in 'nums'. you must decrease the overall operation steps as much as
* possible.
*/
class Solution {
public:
bool search(vector<int> &nums, int target) {
int left = 0, right = nums.size() - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target)
return true;
if (nums[left] == nums[mid] && nums[right] == nums[mid])
left++, right--;
else if (nums[left] <= nums[mid]) {
if (nums[left] <= target && nums[mid] > target)
right = mid - 1;
else
left = mid + 1;
} else {
if (nums[mid] < target && nums[right] >= target)
left = mid + 1;
else
right = mid - 1;
}
}
return false;
}
};
int main() {
Solution obj;
vector<int> nums = {2, 5, 6, 0, 0, 1, 2};
printf("%d\n", obj.search(nums, 0)); // expect: 1
printf("%d\n", obj.search(nums, 3)); // expect: 0
}
|