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. class Solution {
  4. public:
  5.     int search(const vector<int>& nums, int target) {
  6.         int first = 0, last = nums.size();
  7.         while (first != last) {
  8.             const int mid = first  + (last - first) / 2;
  9.             if (nums[mid] == target)
  10.                 return mid;
  11.             if (nums[first] <= nums[mid]) {
  12.                 if (nums[first] <= target && target < nums[mid])
  13.                     last = mid;
  14.                 else
  15.                     first = mid + 1;
  16.             } else {
  17.                 if (nums[mid] < target && target <= nums[last-1])
  18.                     first = mid + 1;
  19.                 else
  20.                     last = mid;
  21.             }
  22.         }
  23.         return -1;
  24.     }
  25. };

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