LeetCode刷题笔录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.

这题也是要用二分法，不然太慢了。不同的是如果把数组从中间平分成两半，只能保证其中一半是完全排好序的。

public class Solution {
public int search(int[] A, int target) {
return search(A, target, 0, A.length - 1);
}

public int search(int[] A, int target, int low, int high){
if(low > high)
return -1;
int mid = (low + high) / 2;
if(A[mid] == target)
return mid;

if(A[low] <= A[mid]){
if(A[low] <= target && target <= A[mid])
return search(A, target, low, mid - 1);
else
return search(A, target, mid + 1, high);
}
else{
if(A[mid] <= target && target <= A[high])
return search(A, target, mid + 1, high);
else
return search(A, target, low, mid - 1);
}
}
}