LeetCode: Search in Rotated Sorted Array 解题报告

Search in Rotated Sorted Array

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.

SOLUTION 1:

使用九章算法经典递归模板：

while (l < r - 1) 可以避免mid与left重合的现象。

其实和一般的二分法搜索没有太多区别。

问题是我们每次要找出正常排序的部分，你只需要比较mid, left，如果它们是正序，就代表左边是

正常排序，而右边存在断开的情况，也就是因为Rotated发生不正常序列。

例如：

4567012 如果我们取mid为7，则左边是正常序列，而右边7012不正常。

然后 我们再将target与正常排序的这边进行比较，如果target在左边，就丢弃右边，反之，丢弃

左边。一次我们可以扔掉一半。和二分搜索一样快。

 1 public class Solution {

 2     public int search(int[] A, int target) {

 3         if (A == null || A.length == 0) {

 4             return -1;

 5         }

 6         

 7         int l = 0;

 8         int r = A.length - 1;

 9         

10         while (l < r - 1) {

11             int mid = l + (r - l) / 2;

12             

13             if (A[mid] == target) {

14                 return mid;

15             }

16             

17             // left side is sorted.

18             // BUG 1: if don't use >= , and use L < r in while loop, than there is some problem.

19             if (A[mid] > A[l]) {

20                 if (target > A[mid] || target < A[l]) {

21                     // move to right;

22                     l = mid + 1;

23                 } else {

24                     r = mid - 1;

25                 }

26             } else {

27                 if (target < A[mid] || target > A[r]) {

28                     // move to left;

29                     r = mid - 1;

30                 } else {

31                     l = mid + 1;

32                 }

33             }

34         }

35         

36         if (A[l] == target) {

37             return l;

38         } else if (A[r] == target) {

39             return r;

40         }

41         

42         return -1;

43     }

44 }

View Code

SOLUTION 2:

注意，如果while 循环使用l <= r来卡，则mid有可能会靠到l这这来，所以当判断是否有序时，我们必须使用<=

总之，这份代码就不需要最后再判断l,r的值。

 1 public int search(int[] A, int target) {

 2         if (A == null || A.length == 0) {

 3             return -1;

 4         }

 5         

 6         int l = 0;

 7         int r = A.length - 1;

 8         

 9         while (l <= r) {

10             int mid = l + (r - l) / 2;

11             

12             if (A[mid] == target) {

13                 return mid;

14             }

15             

16             // left side is sorted.

17             // BUG 1: if don't use >= , and use L < r in while loop, than there is some problem.

18             if (A[mid] >= A[l]) {

19                 if (target > A[mid] || target < A[l]) {

20                     // move to right;

21                     l = mid + 1;

22                 } else {

23                     r = mid - 1;

24                 }

25             } else {

26                 if (target < A[mid] || target > A[r]) {

27                     // move to left;

28                     r = mid - 1;

29                 } else {

30                     l = mid + 1;

31                 }

32             }

33         }

34         

35         return -1;

36     }

View Code

 

FOLLOWUP:

LeetCode 新题: Find Minimum in Rotated Sorted Array 解题 ...

LeetCode 新题: Find Minimum in Rotated Sorted Array II 解 ...

 

代码： GitHub代码链接