Search in Rotated Sorted Array II

描述

Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

分析

允许重复元素，则上一题中如果A[left] <= A[mid],那么[left,mid]为递增序列的假设就不能成立了，比如[1,3,1,1,1]。

既然A[left] <= A[mid]不能确定递增，那就把它拆分成两个条件：

  • 若A[left] < A[mid]，则区间[left,mid]一定递增
  • 若A[left] == A[mid] 确定不了，那就left++，往下看一步即可。

    代码

  1. // Search in Rotated Sorted Array II
  2. // Time Complexity: O(n)，Space Complexity: O(1)
  3. class Solution {
  4. public:
  5.     bool 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 true;
  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 if (nums[first] > nums[mid]) {
  17.                 if (nums[mid] < target && target <= nums[last-1])
  18.                     first = mid + 1;
  19.                 else
  20.                     last = mid;
  21.             } else
  22.                 //skip duplicate one
  23.                 first++;
  24.         }
  25.         return false;
  26.     }
  27. };

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