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     }