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.
(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.
分析:
虽然有序数组折了,但是二分的时候一定有一边是有序的,利用中间元素和右端点元素相比来判断哪一边是有序的。
A[mid] <= A[end], 说明,mid到end之间的元素一定是有序的,
反之,说明,start到mid之间的元素一定是有序的。
判断出来那边有序之后,先判断target是不是在有序的部分里,如果是,就在有序的部分里查找,不是的归为另一类。
public class Solution {
public int search(int[] A, int target) {
return search(A, target, 0, A.length-1);
}
public int search(int[] A, int target, int start, int end){
if(start>end)
return -1;
int mid=(start+end)/2;
if(target == A[mid])
return mid;
if(A[mid] >= A[end]){//左边有序
if(target < A[mid] && target>=A[start])
return search(A, target, start, mid-1);
else
return search(A, target, mid+1, end);
}else{//右边有序
if(target > A[mid] && target<=A[end])
return search(A, target, mid+1, end);
else
return search(A, target, start, mid-1);
}
}
}