Algorithm
Problem Name: 81. Search in Rotated Sorted Array II
problem Link: https://leetcode.com/problems/search-in-rotated-sorted-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.
You must decrease the overall operation steps as much as possible.
Example 1:
Input: nums = [2,5,6,0,0,1,2], target = 0 Output: true
Example 2:
Input: nums = [2,5,6,0,0,1,2], target = 3 Output: false
Constraints:
1 <= nums.length <= 5000-104 <= nums[i] <= 104numsis guaranteed to be rotated at some pivot.-104 <= target <= 104
Code Examples
#1 Code Example with C Programming
Code -
C Programming
bool search(int* nums, int numsSize, int target){
int start, end, mid;
start = 0; end = numsSize - 1;
while (start < = end) {
mid = start + (end - start) / 2;
if (nums[mid] == target) return true;
if (nums[start] == nums[mid]) {
start ++;
} else if (nums[start] < nums[mid]) { // first half are sorted
if (target > nums[mid] || target < nums[start]) {
start = mid + 1;
} else {
end = mid - 1;
}
} else { // second half are sorted
if (target < nums[mid] || target > nums[end]) {
end = mid - 1;
} else {
start = mid + 1;
}
}
}
return false;
}
Input
Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public boolean search(int[] nums, int target) {
int start = 0;
int end = nums.length - 1;
while (start < = end) {
int mid = (start + end) / 2;
if (nums[mid] == target) {
return true;
}
if (nums[mid] < nums[end] || nums[mid] < nums[start]) {
if (target > nums[mid] && target <= nums[end]) {
start = mid + 1;
} else {
end = mid - 1;
}
} else if (nums[mid] > nums[start] || nums[mid] > nums[end]) {
if (target < nums[mid] && target >= nums[start]) {
end = mid - 1;
} else {
start = mid + 1;
}
} else {
end--;
}
}
return false;
}
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
class Solution:
def search(self, nums, target):
l, r, n = 0, len(nums) - 1, len(nums)
while l <= r:
while l + 1 < n and nums[l + 1] == nums[l]:
l += 1
while r > 0 and nums[r] == nums[r - 1]:
r -= 1
mid = (l + r) // 2
if nums[mid] == target:
return True
elif sum((target < nums[l], nums[l] <= nums[mid], nums[mid] < target)) == 2:
l = mid + 1
else:
r = mid - 1
return False
Input
Output
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def search(self, nums, target):
l, r, n = 0, len(nums) - 1, len(nums)
while l <= r:
while l + 1 < n and nums[l + 1] == nums[l]:
l += 1
while r > 0 and nums[r] == nums[r - 1]:
r -= 1
mid = (l + r) // 2
if nums[mid] == target:
return True
elif sum((target < nums[l], nums[l] <= nums[mid], nums[mid] < target)) == 2:
l = mid + 1
else:
r = mid - 1
return False
Input
Output
#5 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _081_SearchInRotatedSortedArray2
{
public bool Search(int[] nums, int target)
{
int lo = 0, hi = nums.Length - 1;
int mid, loValue, hiValue, midValue;
while (lo < = hi)
{
loValue = nums[lo];
hiValue = nums[hi];
if (loValue < hiValue && (target < loValue || target > hiValue))
{
return false;
}
while (lo < hi && nums[lo] == hiValue)
{
lo++;
}
loValue = nums[lo];
mid = lo + (hi - lo) / 2;
midValue = nums[mid];
if (target == midValue) { return true; }
if (loValue < = midValue)
{
if (loValue <= target && target < midValue)
{
hi = mid - 1;
}
else
{
lo = mid + 1;
}
}
else
{
if (target < = hiValue && midValue < target)
{
lo = mid + 1;
}
else
{
hi = mid - 1;
}
}
}
return false;
}
}
}
Input
Output