LEETCODE: Search in Rotated Sorted Array II

原创于 2014-12-26 21:42:45 发布 · 271 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#数组 #LEETCODE #算法

LEETCODE 专栏收录该内容

153 篇文章

订阅专栏

本文探讨了在允许元素重复的旋转排序数组中查找特定目标值的问题。文章提供了一个C++实现示例，该算法考虑了重复元素如何影响时间复杂度，并通过调整二分查找策略来解决这一挑战。

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) {
        int left = 0;
        int right = n - 1;
        while(left <= right) {
            int mid = (left + right) / 2;
            if(A[mid] == target) return true;
            if(A[left] == A[mid]) {
                left ++;
                continue;
            }
            if(A[right] == A[mid]) {
                right --;
                continue;
            }
            if(A[left] < A[mid] ) {
				if(target < A[mid]  && target >= A[left])
					right = mid - 1;
				else
					left = mid + 1;
            }
            else { 
				if(target <= A[right] && target > A[mid]) 
					left = mid + 1;
				else
					right = mid - 1;
            }
        }
        return false;
    }
};