Search in Rotated Sorted Array

描述

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.

分析

一个有序数组被循环右移，只可能有一下两种情况：

  1.    7 │
  2.  6   │
  3. ─────┼───────────
  4.      │         5
  5.      │       4
  6.      │     3
  7.      │   2
  8.      │ 1
  1.          7 │
  2.        6   │
  3.      5     │
  4.    4       │
  5.  3         │
  6. ───────────┼───────────      
  7.            │   2
  8.            │ 1

本题依旧可以用二分查找，难度主要在于左右边界的确定。仔细观察上面两幅图，我们可以得出如下结论：

如果A[left] <= A[mid],那么[left,mid] 一定为单调递增序列。

代码

  1. // Search in Rotated Sorted Array
  2. // Time Complexity: O(log n)，Space Complexity: O(1)
  3. public class Solution {
  4.     public int search(int[] nums, int target) {
  5.         int first = 0, last = nums.length;
  6.         while (first != last) {
  7.             final int mid = first  + (last - first) / 2;
  8.             if (nums[mid] == target)
  9.                 return mid;
  10.             if (nums[first] <= nums[mid]) {
  11.                 if (nums[first] <= target && target < nums[mid])
  12.                     last = mid;
  13.                 else
  14.                     first = mid + 1;
  15.             } else {
  16.                 if (nums[mid] < target && target <= nums[last-1])
  17.                     first = mid + 1;
  18.                 else
  19.                     last = mid;
  20.             }
  21.         }
  22.         return -1;
  23.     }
  24. };

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原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/search/search-in-rotated-sorted-array.html