LeetCode – Search in Rotated Sorted Array (Java)
Suppose a sorted array 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.
Analysis
In order to use binary search on the rotated sorted array, we need to determine how to update the left and right pointers. There are two major cases as shown below:
Once the two cases are identified, the problem is straightforward to solve. We only need to check if the target element is in the sorted side, and based on that move left or right pointers.
Java Solution 1 Recusive
public int search(int[] nums, int target) { return binarySearch(nums, 0, nums.length1, target); } public int binarySearch(int[] nums, int left, int right, int target){ if(left>right) return 1; int mid = left + (rightleft)/2; if(target == nums[mid]) return mid; if(nums[left] <= nums[mid]){ if(nums[left]<=target && target<nums[mid]){ return binarySearch(nums,left, mid1, target); }else{ return binarySearch(nums, mid+1, right, target); } }else { if(nums[mid]<target&& target<=nums[right]){ return binarySearch(nums,mid+1, right, target); }else{ return binarySearch(nums, left, mid1, target); } } } 
Java Solution 2  Iterative
public int search(int[] nums, int target) { int left = 0; int right= nums.length1; while(left<=right){ int mid = left + (rightleft)/2; if(target==nums[mid]) return mid; if(nums[left]<=nums[mid]){ if(nums[left]<=target&& target<nums[mid]){ right=mid1; }else{ left=mid+1; } }else{ if(nums[mid]<target&& target<=nums[right]){ left=mid+1; }else{ right=mid1; } } } return 1; } 
<pre><code> String foo = "bar"; </code></pre>

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