Search in Rotated Sorted Array

最新推荐文章于 2024-05-03 20:14:24 发布

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最新推荐文章于 2024-05-03 20:14:24 发布

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分类专栏： Leedcode

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本文介绍了一种在旋转过的有序数组中查找特定元素的算法。针对两种情况：无重复元素和有重复元素，分别提出了二分查找法的解决方案，并讨论了算法的时间复杂度。

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初阶。

Suppose a sorted array 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).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.

进阶。

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.

对于初阶。思路，采用二分查找。如果中间数大于最左边的数，那么左半部分必然有序，否则右半部分必然有序。通过目标数与有序部分的最大最小数比较，可以判断目标数到底属于哪一个部分。时间复杂度O(logn),空间复杂度O(1).

class Solution{
public:
	int search(int A[],int n,int target){
		int first = 0,last = n;
		while(first != last){
			int mid = (first + last)/2;
			if (A[mid] == target)
				return mid;
			if(A[first] <= A[mid]){
				if (A[first] <= target && target < A[mid])
					last = mid;
				else
					first = mid + 1;
			}else{
				if (A[mid] < target && target <= A[last-1])
					first = mid + 1;
				else
					last = mid;				
			}

		}
		return -1;
	}
};

对于进阶。思路，和初阶的区别在于。如果A[m]>=A[l],不能保证[l,m]是有序的。比如[2,4,2,2,2].这种情况下可以拆分为两个条件，1）A[m]>a[l],则[l,m]必然有序；2）如果A[m]==a[l],那么就l++,往下看。

int search_2(int A[],int n,int target){
		int first = 0,last =n;
		while(first != last)
		{
			int mid = (first + last)/2;
			if(A[mid] == target)
				return mid;
			if(A[first] < A[mid]){
				if(A[first]<=target && target < A[mid])
					last = mid;
				else
					first = mid + 1;
			}else if(A[first] > A[mid]){
				if(A[mid] < target && target <= A[last-1])
					first = mid + 1;
				else
					last = mid;
			}else
				//skip duplicate one
				first ++;
		}
		return -1;
	}