81. Search in Rotated Sorted Array II
Last updated
Last updated
/**
* Time complexity : O(n)
* Space complexity : O(1)
*/
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;
}
}/**
* Time complexity : O(logN)
* Space complexity : O(1)
*/
class Solution {
public boolean search(int[] nums, int target) {
int left = 0, right = nums.length-1;
while(left <= right) {
int mid = left + (right-left)/2;
if(target == nums[mid]) return true;
if(nums[left] < nums[mid]) {
if(target >= nums[left] && target <= nums[mid]) {
right = mid - 1;
} else {
left = mid + 1;
}
} else if(nums[left] > nums[mid]) {
if(target >= nums[mid] && target <= nums[right]) {
left = mid + 1;
} else {
right = mid - 1;
}
} else {
left++;
}
}
return false;
}
}