LeetCode刷题【Array】 Search in Rotated Sorted Array II

题目：

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

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

Suppose an array sorted in ascending order 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).

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

The array may contain duplicates.

解决方法一：主要是考虑到重复数情况 如[1,1,1,1,2,1]

public class Solution {    public boolean search(int[] nums, int target) {         int mid =0;          int start =0;          int end = nums.length-1;          boolean ret = false;          while(start<=end){              mid =(start+end)/2;              if(target==nums[mid]){                  ret = true;                  break;              }             if(start==end) break;            if(nums[mid]>nums[end]){                  end = target>=nums[start]&&target<nums[mid]? mid-1:end;                  start = target>=nums[start]&&target<nums[mid]?start:mid+1;              }              else if(nums[mid]<nums[start]){                  start = target<=nums[end]&&target>nums[mid]? mid+1:start;                  end = target<=nums[end]&&target>nums[mid]?end:mid-1;              }else if(nums[mid]==nums[end]&&nums[mid]==nums[start]){                start++;                end--;            } else{                  start = target<=nums[mid]? start:mid+1;                  end = target<=nums[mid]? mid-1:end;              }          }          return ret;      }}

参考：

【1】https://leetcode.com/