Problem
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,0,1,2,2,5,6]
might become [2,5,6,0,0,1,2]
).
You are given a target value to search. If found in the array return true
, otherwise return false
.
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
Follow up:
- This is a follow up problem to Search in Rotated Sorted Array, where
nums
may contain duplicates. - Would this affect the run-time complexity? How and why?
Solution
/**
* @param {number[]} nums
* @param {number} target
* @return {boolean}
*/
var search = function(nums, target) {
var left = 0;
var right = nums.length - 1;
var mid = 0;
while (left <= right) {
mid = Math.floor((left + right) / 2);
if (nums[mid] === target) return true;
if (nums[mid] > nums[left]) {
if (nums[left] <= target && target < nums[mid]) {
right = mid - 1;
} else {
left = mid + 1;
}
} else if (nums[mid] < nums[left]) {
if (nums[mid] < target && target <= nums[right]) {
left = mid + 1;
} else {
right = mid - 1;
}
} else {
left++;
}
}
return false;
};
Explain:
see Search in Rotated Sorted Array.
- 判断哪边是有序的
- 判断
target
在有序的那边还是无序的那边
注意重复数字的情况下,只能一个个移动,因为没法判断在哪边。这样算法最坏的情况就是 O(n)
了。
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).