首页 > 代码库 > 【leetcode刷题笔记】Search in Rotated Sorted Array II

【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 }