[LeetCode] 33. Search in Rotated Sorted Array

题：https://leetcode.com/problems/search-in-rotated-sorted-array/description/

问题

Suppose an array sorted in ascending order 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

1.用二分查找的方式去求解，但亮点在于可以确定旋转后mid的位置realmid。
2.求解旋转点时，因为旋转点即为数字串中最小值的位子。问题变为求最小值的位置，利用数组的部分有序，实行二分法查找（重要）。

直接放别人做的源码，上面有很详细地注释。
源码

class Solution {public:    int search(int A[], int n, int target) {        int lo=0,hi=n-1;        // find the index of the smallest value using binary search.        // Loop will terminate since mid < hi, and lo or hi will shrink by at least 1.        // Proof by contradiction that mid < hi: if mid==hi, then lo==hi and loop would have been terminated.        while(lo<hi){            int mid=(lo+hi)/2;            if(A[mid]>A[hi]) lo=mid+1;            else hi=mid;        }        // lo==hi is the index of the smallest value and also the number of places rotated.        int rot=lo;        lo=0;hi=n-1;        // The usual binary search and accounting for rotation.        while(lo<=hi){            int mid=(lo+hi)/2;            int realmid=(mid+rot)%n;            if(A[realmid]==target)return realmid;            if(A[realmid]<target)lo=mid+1;            else hi=mid-1;        }        return -1;    }};