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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;
}
leetcode之Search in Rotated Sorted Array
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原文地址：http://www.cnblogs.com/gracyandjohn/p/4403464.html
