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.
思路:Binary Search ,但是此题的规则要稍作修改,要判断左半边是否有序。
class Solution {
public:
int search(int A[], int n, int target) {
if(n==0)
return -1;
int left = 0;
int right = n-1;
int middle = 0;
while(left<=right){
middle = left+(right-left)/2;
if(A[middle]==target)
return middle;
if(A[middle]>=A[left]){
if(A[left]<=target&&target<A[middle]){
right = middle-1;
}
else
left = middle+1;
}
else{
if(A[middle]<target&&target<=A[right]){
left = middle+1;
}
else
right = middle-1;
}
}
return -1;
}
};