Search in Rotated Sorted Array

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本文介绍了一种用于在已排序并旋转的数组中搜索目标值的算法。该算法利用了数组的有序性质和旋转特性，通过二分查找进行优化，确保在O(log n)的时间复杂度内找到目标值或返回-1。详细步骤包括确定数组左侧或右侧是否有序，然后判断目标值是否位于相应范围内。

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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:

1. if target==A[mid], return.

2. if left is in order, then determine whether target fits into that specific range or not.

3. if right is in order, then similarly, determine whether target fits into the range or not.

public class Solution {
    public int search(int[] A, int target) {
        int low=0, high=A.length-1;
        
        while(low <= high) {
            int mid = (low+high)/2;
            if(target==A[mid]) return mid;
            if(A[mid] >= A[low]) {   // left in order
                if(target<A[mid] && target>=A[low]) high=mid-1;
                else low=mid+1;
            }
            if(A[mid] <= A[high]) {  // right in order
                if(target>A[mid] && target<=A[high]) low=mid+1;
                else high=mid-1;
            }
        }
        
        return -1;
    }
}