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.
Your algorithm's runtime complexity must be in the order of O(log n).
Example 1:
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2], target = 3 Output: -1
法1：将数组一分为二，其中一定有一个是有序的，另一个可能是有序，也能是部分有序。此时有序部分用二分法查找。无序部分再一分为二，其中一个一定有序，另一个可能有序，可能无序。就这样循环.
class Solution {
public:
int search(vector<int>& nums, int target) {
if(nums.size()==0) return -1;
return bt_search(nums, 0,nums.size()-1,target);
}
int bt_search(vector<int>& a, int low, int high, int target) {
int mid = low + (high - low) / 2;
if(a[low]==target) return low;
if(a[high]==target) return high;
if(low>=high) return -1;
if (a[low] < a[mid]) {
if (a[low] <target && target< a[mid]) {
return bt_search(a, low+1, mid-1, target);
}
else {
return bt_search(a,mid, high, target);
}
} else {
if (a[mid] <target && target <a[high]) {
return bt_search(a, mid+1, high-1, target);
}
else {
return bt_search(a,low, mid, target);
}
}
return -1;
}
};
从左向右，如果左边的点比右边的点小，说明这两个点之间是有序的。
如果左边的点比右边的点大，说明中间有个旋转点，所以一分为二后，肯定有一半是有序的。所以还可以用二分法。
不过先要判断左边有序还是右边有序，如果左边有序，则直接将目标与左边的边界比较，就知道目标在不在左边，
如果不在左边肯定在右边。
1 class Solution { 2 public int search(int[] a, int target) { 3 int n = a.length; 4 int lo = 0; 5 int hi = n - 1; 6 while(lo<=hi){ 7 int mid = lo+(hi-lo)/2; 8 if(a[mid]== target) 9 return mid; 10 11 if(a[lo]<=a[mid]){//左半边有序 12 if(a[lo]<=target && target<=a[mid])//目标值在左半边 13 hi = mid - 1; 14 else 15 lo = mid + 1; 16 } 17 else{//右半边有序 18 if(a[mid]<=target && target<=a[hi]) 19 lo = mid + 1; 20 else 21 hi = mid - 1; 22 } 23 } 24 return -1; 25 26 } 27 }