0033_Search in Rotated Sorted Array
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Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
JAVA
由于不是完全的升序,因此不能直接套用二分查找。对于数组4 5 6 7 0 1 2,将[4,7]定义为大段,将[0,2]定义为小段。最初想直接找到大段和小段的分界线,然后再套用二分查找,但是发现定位这个分界点需要的时间和直接查找target差不多了。随后执行使用二分查找的思想,然后对不同的情况进行判断。刚开始的时候想使用一个对于大段和小段都可直接复用的比较方法,在纠结了很久之后发现行不通。于是直接先定位nums[mid]是大段还是小段,然后再对下一个查找范围进行判断。由此可见,虽然我一直相信懒是人类进步的阶梯,所以凡事能写一个函数的我绝不写两个,但是以此题为例,我发现有时候想出一个通用的函数来做一个非常简单的事,还不如直接多写一下函数或判断开发效率更高。。。
时间效率在前1/3。
public class Solution { public int search(int[] nums, int target) { int left = 0; int right = nums.length - 1; int mid = 0; while(left <= right){ mid = (left + right) / 2; if(target == nums[mid]){ return mid; }else if(target == nums[right]){ return right; }else if(target == nums[left]){ return left; } else if(nums[mid] > nums[right]){ if(target > nums[mid]){ left = mid + 1; }else{ if(target < nums[left]){ left = mid + 1; }else{ right = mid - 1; } } }else{ if(target > nums[mid]){ if(target > nums[right]){ right = mid - 1; }else{ left = mid + 1; } }else{ right = mid - 1; } } } return -1; }}
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