33. Search in Rotated Sorted Array

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.

在旋转了的有序数组里面查找 

在正常的有序数组中 我们可以使用二分查找 是因为整个数组是有序的 把数组从中间分开 仍然是两个有序数组

旋转了的有序数组 在旋转点的前后 分别是有序数组 数组从中间分开 那么至少有一个是有序数组 对于这部分 我们仍然可以采用二分查找

public int search(int[] nums, int target) {
int start = 0, end = nums.length-1;
while (start <= end) {
int mid = (start+end) >>> 1;
if (nums[mid] == target) {
return mid;
} else if (nums[mid] >= nums[start]) {
if (target >=nums[start] && target < nums[mid]) {
end = mid-1;
} else start = mid+1;
} else {
if (target > nums[mid] && target <= nums[end]) {
start = mid+1;
} else end = mid-1;
}
}
return -1;
}