[LeetCode刷题记录]Search in Rotated Sorted Array

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本文介绍了一种高效的方法来搜索在未知旋转点的排序数组中目标值的位置，通过判断中间点左右两侧的有序性进行搜索。

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.

思路：关键是判断middle点左右半边那一侧是有序的。（3种情况，左右一侧有序，左右都有序）

比如如果左侧有序，而target>A[left],target<A[right]，说明target在左侧，就在左侧搜索。

如果target不在该范围内，就在右侧搜索。

右侧有序同理。

class Solution {
public:
    int search(int A[], int n, int target) {
        if(n==0)
            return -1;
        int left = 0;
        int right = n-1;
        int middle = 0;
        while(left<=right){
            middle = left+(right-left)/2;
            if(A[middle]==target)
                return middle;
            if(A[middle]>=A[left]){
                if(A[left]<=target&&target<A[middle]){
                    right = middle-1;
                }
                else
                    left = middle+1;
            }
            else{
                if(A[middle]<target&&target<=A[right]){
                    left = middle+1;
                }
                else
                    right = middle-1;
            }
        }
        return -1;
    }
};