[LeetCode刷题记录]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.

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;
    }
};