leetCode 81.Search in Rotated Sorted Array II (旋转数组的搜索II) 解题思路和方法
Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
Write a function to determine if a given target is in the array.
思路:此题在解的时候,才发现Search in Rotated Sorted Array的时候想复杂了,事实上仅仅须要遍历搜索就可以,线性时间。
代码例如以下:
public 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; } }