leetCode 81.Search in Rotated Sorted Array II （旋转数组的搜寻II） 解题思路和方法

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，其实只需要遍历搜索即可，线性时间。

代码如下：

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;
    }
}