[leetcode]33. Search in Rotated Sorted Array旋转过有序数组的查找问题

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.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

题目

一个有序数组，可能进行了循环移位，在里面进行查找。

思路

1. We can observe that at least one half of given array is sorted.

2. Find such sorted array part and do binary search

代码

 1 class Solution {
 2     public int search(int[] nums, int target) { 
 3         int n = nums.length ;
 4         int left = 0, right = n-1;
 5         while(left<=right) {
 6             int mid = left + (right-left)/2;
 7             if(nums[mid]==target) return mid;
 8 
 9             if(nums[mid]<nums[right]) { // right half sorted
10                 if(target>nums[mid] && target<=nums[right])
11                     left = mid+1;
12                 else
13                     right = mid-1;
14             }
15             else if(nums[mid]>nums[right]) {  // left half sorted
16                 if(target>=nums[left] && target<nums[mid])
17                     right = mid-1;
18                 else
19                     left = mid+1;
20             }
21             else {
22                 right--;
23             }
24         }
25         return -1;
26     }
27 }