LeetCode刷题【Array】 Search in Rotated Sorted Array II

最新推荐文章于 2025-07-29 14:24:32 发布

原创最新推荐文章于 2025-07-29 14:24:32 发布 · 355 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Array #leetcode

LeetCode 专栏收录该内容

49 篇文章

订阅专栏

针对含有重复元素的旋转排序数组，本文介绍了一种查找特定目标值的有效算法。该方法通过调整二分查找策略来处理重复元素带来的挑战，确保了算法在复杂情况下的正确性和效率。

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/