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. public class Solution {
  4.     public boolean 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 true;
  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 if (nums[first] > nums[mid]) {
  16.                 if (nums[mid] < target && target <= nums[last-1])
  17.                     first = mid + 1;
  18.                 else
  19.                     last = mid;
  20.             } else
  21.                 //skip duplicate one
  22.                 first++;
  23.         }
  24.         return false;
  25.     }
  26. };

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