LeetCode 33.Search in Rotated Sorted Array

题目：

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.

分析与解答：

这个数组的特点是，如果pivot在区间(a,b]内的话，那么a一定是大于等于b的。依然是用二分搜索，只是在边界条件的判断上根据数组的特点做一些改动。

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