【leetcode刷题笔记】Search in Rotated Sorted Array II

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.

题解：如果没有重复的元素，那么就可以根据target是否在某一半而扔掉另外一半。但是如果有重复的元素，就有可能不知道往哪边跳转：

例如A = {1,3,3,3,3,3}，经过变换后得到数组{3,1,3,3,3,3}，此时A[mid] = 3 = A[left] = A[right]，如果target = 1，两边都不能扔，所以不能用二分的方法。

直接用遍历O(n)的方法也可以AC：

 1 public class Solution {
 2     public boolean search(int[] A, int target) {
 3         if(A == null || A.length == 0)
 4             return false;
 5         for(int i = 0;i < A.length;i++)
 6             if(target == A[i])
 7                 return true;
 8         return false;
 9     }
10 }

【leetcode刷题笔记】Search in Rotated Sorted Array II,布布扣,bubuko.com

【leetcode刷题笔记】Search in Rotated Sorted Array II

原文：http://www.cnblogs.com/sunshineatnoon/p/3857995.html