[LeetCode]Search in Rotated Sorted Array II

最新推荐文章于 2020-07-22 10:43:35 发布

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最新推荐文章于 2020-07-22 10:43:35 发布

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分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/sunbaigui/article/details/8981288

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本文介绍了一种在旋转数组中查找特定元素的搜索算法。通过分析旋转数组的特点，该算法能够高效地确定目标值是否存在。文章提供了两种实现方式，并详细解释了每一步的逻辑判断过程。

class Solution {
//classified discussion
//1. Based on the property of rotated array, there may or may not have one sorted sequence 
//when one sequence is divided into two parts  
//2. make decision under all these cases
public:
	bool search(int A[], int n, int target) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		return BinarySearch(A, n, target);
	}


	bool BinarySearch( int* A, int n, int target ) 
	{
		//throw std::exception("The method or operation is not implemented.");
		int l = 0;
		int r = n-1;
		while (l <= r)
		{
			int m = l+(r-l)/2;
			if(A[m] == target) return true;
			if (A[l] < A[m])//left subsequence sorted
			{
				if(A[l] <= target && target < A[m])
					r = m-1;
				else l = m+1;
			}
			else if (A[m] < A[r])
			{
				if(A[m] < target && target <= A[r])
					l = m+1;
				else r = m-1;
			}
			else if (A[l] == A[m])//A[m] is not the target, so remove en element equal to A[m] is safe
				l++;
			else if(A[m] == A[r])//ditto
				r--;
		}
		return false;
	}
};

second time

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