LeetCode: Search in Rotated Sorted Array

Title：

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：还是二分搜索，但是对于一个中间值，如何处理是左加还是右加呢？考虑下，旋转之后的数组，中间值会将原来的数组划分为左右两个数组，其中一个是有序的。所以我们需要做的事就是判断那边是有序的。如何判断有序呢？只要将中间值和最左边的值比较，如果大于，则左边是有序，如果小于，则右边有序，如果相等，则有重复元素只能一个个处理。

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