81. Search in Rotated Sorted Array II**

最新推荐文章于 2023-11-12 22:36:10 发布

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最新推荐文章于 2023-11-12 22:36:10 发布

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分类专栏： leetcode Java Binary Search

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本文介绍了一种在含有重复元素的旋转有序数组中查找特定目标值的方法。文章提供了一个Java实现示例，通过调整二分搜索策略来处理数组旋转及重复元素带来的挑战。

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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.

陷阱：1 3 1 1 1 ; 3 1

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

总结：注意转折点