[leetcode] Search in Rotated Sorted Array II

最新推荐文章于 2020-08-30 11:00:31 发布

原创最新推荐文章于 2020-08-30 11:00:31 发布 · 549 阅读

0 ·

CC 4.0 BY-SA版权

算法题专栏收录该内容

107 篇文章

订阅专栏

本文探讨了在允许重复元素的情况下，如何优化搜索旋转排序数组的算法，并分析了这种变化对运行时间复杂度的影响。通过实现并比较不同策略，揭示了搜索效率的变化路径。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

class Solution {
public:
    bool search(int A[], int n, int target) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		int left,right,middle;
		left=0,right=n-1;
		while(left<=right){
			middle=((right-left)>>1)+left;
			if(A[middle]==target)
				return true;
			if(A[left]<A[middle]){//left is sorted
				if(A[left]<=target && target<A[middle])
					right=middle-1;
				else
					left=middle+1;

			}else if(A[left]>A[middle]){//right is sorted
				if(A[middle]<target && target<=A[right])
					left=middle+1;
				else
					right=middle-1;
			}
			else
				left++;
		}
		return false;
	}
};