Search in Rotated Sorted Array Medium
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.
public int search(int[] A, int target) {
int lo = 0;
int hi = A.length - 1;
if (hi < 0)
return -1;
while (lo < hi) {
int mid = (lo + hi) / 2;
if (A[mid] == target)
return mid;
if (A[lo] <= A[mid]) {
if (target >= A[lo] && target < A[mid])
hi = mid - 1;
else
lo = mid + 1;
} else {
if (target > A[mid] && target <= A[hi])
lo = mid + 1;
else
hi = mid - 1;
}
}
return A[lo] == target ? lo : -1;
}
思路:二分查找其实并不需要整个数组有序,其中有一半有序即可。一个变形的二分查找,判断条件:先和0点比,可以判断是怎么Rotate的,然后再判断应该落在哪个区间。O(lgn)