Search In Rotated Array II : 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 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
, or false
if it is not in nums
.
Example :
Input: nums = [2,5,6,0,0,1,2], target = 0 Output: true
Table of Contents
Search In Rotated Array II Problem Solution
Problem Solution In Python
class Solution:
def search(self, nums: List[int], target: int) -> bool:
pivot=0
for ar in range(len(nums)-1):
if nums[ar]>nums[ar+1]:
pivot=ar+1
break
nums = nums[pivot:]+nums[0:pivot]
left=0
right=len(nums)-1
while left<=right:
mid = int((left+right)/2)
if nums[mid]==target:
return True
elif nums[mid]>target:
right=mid-1
else:
left=mid+1
return False
Problem Solution In Java
class Solution {
public boolean search(int[] nums, int target)
{
for(int i=0;i<nums.length;i++)
{
if(nums[i]==target)
return true;
}
return false;
}
}
Problem Solution In C++
class Solution {
public:
bool search(const vector<int>& nums, int target) {
for (const auto& i : nums) {
if (i == target) {
return true;
}
}
return false;
}
};
Problem Solution In C
bool search(int* nums, int numsSize, int target){
int n = 0;
while (n < numsSize) {
if (nums[n] < target) {
n++;
} else if (nums[n] == target) {
return true;
} else {
break;
}
}
if (n != 0) {
return false;
} else {
int end = numsSize - 1;
while (end > n) {
if (nums[end] > target) {
end--;
}else if (nums[end] == target) {
return true;
} else {
return false;
}
}
}
return false;
}