LeetCode 33 Search in Rotated Sorted Array 巧用递归二分法查找改造

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.
解题思路：正常人都能想到一个解法，遍历所有值，然后返回。这个做法虽然也能accepted，但是违背了出题者的本意。通过一番思考和上网查阅后，学到一个算法，感谢ljiabin大神。他提供的一个求解图让我豁然开朗。



通过这个图，当只考虑升序的时候，基本只有这三种情况，我们只要把所有的情景都能够利用二分法处理，那么这道题很容易。三种情况的区分主要是mid(中点)处在不同位置的结果，正常的二分法则按照图一来处理，其余两种的不同便是mid的不同。通过left
right mid 三个中点的不同值来区分这三种情况。在处理返回值的时候，这里有一个简易的做法，递归的结束都在函数的开始的时候处理，这样的做法减少了代码的冗余。

代码如下：

public int search(int[] nums, int target) {
int len = nums.length-1;

return binarySearch(nums, 0, len, target);
}
public int binarySearch(int[] nums, int left, int right, int target){
int mid = (left+right)/2;
if(left>right)return -1;
if(nums[left]==target)return left;
if(nums[mid]==target)return mid;
if(nums[right]==target)return right;
if(nums[left]<nums[right]){
if(target<nums[left]||target>nums[right]){
return -1;
}else if(target>nums[mid]){
return binarySearch(nums, mid+1, right-1, target);
}else{
return binarySearch(nums, left+1, mid-1, target);
}
}else if(nums[left]<nums[mid]){
if(target>nums[left]&&target<nums[mid]){
return binarySearch(nums, left+1, mid-1, target);
}else{
return binarySearch(nums, mid+1, right-1, target);
}
}else{
if(target<nums[right]&&target>nums[mid]){
return binarySearch(nums, mid+1, right-1, target);
}else{
return binarySearch(nums, left+1, mid-1, target);
}
}
}

参考文章：
http://blog.csdn.net/ljiabin/article/details/40453607
该文章中还提出了一个更简洁的算法：

public class Solution {
public int search(int[] A, int target) {
int l = 0;
int r = A.length - 1;
while (l <= r) {
int mid = (l + r) / 2;
if (target == A[mid]) return mid;
if (A[l] <= A[r]) {
if (target < A[mid]) r = mid - 1;
else l = mid + 1;
} else if (A[l] <= A[mid]) {
if (target > A[mid] || target < A[l]) l = mid + 1;
else r = mid - 1;
} else {
if (target < A[mid] || target > A[r]) r = mid - 1;
else l = mid + 1;
}
}
return -1;
}
}