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.
binary search
但是要分下情况。。。 left rotated, right rotated
4567012 是right rotated 小于一半的跑右面, 或者67012345 ,left rotated
那么可以用中间来判断是左右rotate. 只需要判断好了之后用二分法去search
class Solution {
public:
int search(int A[], int n, int target) {
int l=0, r=n-1;
while(l<=r){
int m=(l+r)/2;
if (A[m]==target)
return m;
if (A[m]>=A[l]){
if (target>=A[l] && target<A[m]){
r=m-1;
} else{
l=m+1;
}
} else{
if (target>A[m] && target <= A[r]){
l=m+1;
}else{
r=m-1;
}
}
}
return -1;
}
};
http://fisherlei.blogspot.com/2013/01/leetcode-search-in-rotated-sorted-array.html