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。

 public class Solution {
     public int search(int[] A, int target) {
         if (A == null || A.length == 0) {
             return -1;
         }

         int l = 0;
         int r = A.length - 1;

         while (l < r - 1) {
             int mid = l + (r - l) / 2;

             if (A[mid] == target) {
                 return mid;
             }

             // left side is sorted.
             // BUG 1: if don't use >= , and use L < r in while loop, than there is some problem.
             if (A[mid] > A[l]) {
                 if (target > A[mid] || target < A[l]) {
                     // move to right;
                     l = mid + 1;
                 } else {
                     r = mid - 1;
                 }
             } else {
                 if (target < A[mid] || target > A[r]) {
                     // move to left;
                     r = mid - 1;
                 } else {
                     l = mid + 1;
                 }
             }
         }

         if (A[l] == target) {
             return l;
         } else if (A[r] == target) {
             return r;
         }

         return -1;
     }
 }

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。

 public int search(int[] A, int target) {
         if (A == null || A.length == 0) {
             return -1;
         }

         int l = 0;
         int r = A.length - 1;

         while (l <= r) {
             int mid = l + (r - l) / 2;

             if (A[mid] == target) {
                 return mid;
             }

             // left side is sorted.
             // BUG 1: if don't use >= , and use L < r in while loop, than there is some problem.
             if (A[mid] >= A[l]) {
                 if (target > A[mid] || target < A[l]) {
                     // move to right;
                     l = mid + 1;
                 } else {
                     r = mid - 1;
                 }
             } else {
                 if (target < A[mid] || target > A[r]) {
                     // move to left;
                     r = mid - 1;
                 } else {
                     l = mid + 1;
                 }
             }
         }

         return -1;
     }

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