Leetcode -- Search in Rotated Sorted Array

问题链接：https://oj.leetcode.com/problems/search-in-rotated-sorted-array/

问题描述：

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.

问题API：public int search(int[] A, int target)

问题分析：这题其实就是典型的二分运用。这个题目再没有重复的情况下是logN， 只需要判断rotate之后中间分界后一边是rotate过后变成unsorted的样子，另一边依旧是sorted的递增的样子，所以判断好递增是哪一边之后，其实就跟Binary Search没有俩样了。代码如下：

    public int search(int[] A, int target) {        if(A == null)            return -1;        int mid, head = 0, tail = A.length - 1;        while(head <= tail){            mid = (head + tail) / 2;            if(target == A[mid])                return mid;            else if(A[mid] < A[tail]){//这就是判断哪边是递增的一边，利用binary search的思想判断target是否在递增的一边，不在的话肯定就是在非递增的一边了                if(target > A[mid] && target <= A[tail])                    head = mid + 1;                else                    tail = mid - 1;            }else{                if(target < A[mid] && target >= A[head])//同上                    tail = mid - 1;                else                    head = mid + 1;            }        }        return -1;    }