81. Search in Rotated Sorted Array II

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).

You are given a target value to search. If found in the array return true, otherwise return false.

Example 1:

Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true

Example 2:

Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false

Follow up:

• This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
• Would this affect the run-time complexity? How and why?

• class Solution {
      public boolean search(int[] nums, int target) {
          int n = nums.length;
          int l = 0, r = n - 1;
          while(l <= r){
              int mid = l + (r-l)/2;
              if(nums[mid] == target) return true;
              if(nums[mid] < nums[r]){//大条件，说明mid到r是升序
                  if(target > nums[mid] && target <= nums[r]) l = mid+1;//需要同时满足这两个条件
                  else r = mid-1;
              }
              else if(nums[mid] > nums[r]){
                  if(target < nums[mid] && target >= nums[l]) r = mid - 1;
                  else l = mid + 1;
              }
              else r--;//中间值等于最右值时，目标值既可以在左边又可以在右边，既然我们比较的是r，那就r--
          }
          return false;
      }
  }

  https://www.cnblogs.com/grandyang/p/4325840.html