leetcode之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.

这道题和剑指offer上面试题8旋转数组的最小数字的思想差不多。这可以看作是二分查找的变种。

先找到最小 元素，然后将原来数组分成两组，分别查找。但是每条通

将代码附在下面

 public int search(int[] A, int target) {
        int left = 0, right = A.length - 1;
		while (left <= right) {
			int mid = (left + right) / 2;
			if (A[mid] == target)
				return mid;
			if (A[left] < A[mid]) {
				if (target <= A[mid] && target >= A[left])
					right = mid - 1;
				else
					left = mid + 1;
			} else if (A[left] > A[mid]) {
				if (target >= A[left] || target <= A[mid])
					right = mid - 1;
				else
					left = mid + 1;
			} else
				left++;
		}
		return -1;
    }