LeetCode OJ - Search in Rotated Sorted Array

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.

Java Solution 1- Recusive

public int search(int[] nums, int target) {
return binarySearch(nums, 0, nums.length-1, target);
}

public int binarySearch(int[] nums, int left, int right, int target){
if(left>right)
return -1;

int mid = left + (right-left)/2;

if(target == nums[mid])
return mid;

if(nums[left] <= nums[mid]){
if(nums[left]<=target && target<nums[mid]){
return binarySearch(nums,left, mid-1, 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, mid-1, target);
}
}
}
Java Solution 2 - Iterative

<pre name="code" class="java">public int search(int[] nums, int target) {
int left = 0;
int right= nums.length-1;

while(left<=right){
int mid = left + (right-left)/2;
if(target==nums[mid])
return mid;

if(nums[left]<=nums[mid]){
if(nums[left]<=target&& target<nums[mid]){
right=mid-1;
}else{
left=mid+1;
}
}else{
if(nums[mid]<target&& target<=nums[right]){
left=mid+1;
}else{
right=mid-1;
}
}
}

return -1;
}

转载自： http://www.programcreek.com/2014/06/leetcode-search-in-rotated-sorted-array-java/

Runtime 都是1ms 就不知道递归与迭代方法的时间复杂度和空间复杂度一样不.....