[LeetCode]Search in Rotated Sorted Array

最新推荐文章于 2024-05-03 20:14:24 发布

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本文提供了一种在旋转排序数组中查找特定元素的有效算法。通过修改二分查找法来适应旋转数组的特点，该方法能在O(log n)时间内找到目标值。文章详细介绍了如何判断数组的有序部分并据此调整搜索范围。

class Solution {
//observation is the key, try to solve it by modifying binary search
public:
	int search(int A[], int n, int target) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		int l = 0;
		int r = n-1;
		while (l <= r)
		{
			int mid = l+(r-l)/2;
			if(A[mid] == target) return mid;
			if (A[l] <= A[mid])//it is sorted in [l,mid], "=" contains an edge case:3, 1
			{
				if(A[mid] > target && A[l] <= target)
					r = mid-1;
				else 
					l = mid+1;
			}
			else//[mid,r] must be sorted, if [l,mid] not sorted
			{
				if(A[mid] < target && A[r] >= target)
					l = mid+1;
				else 
					r = mid-1;
			}
		}
		return -1;
	}
};

second time

class Solution {
public:
    int 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[mid] >= A[left])//left side sorted, including mid,
            {
                if(A[left] <= target && target <= A[mid]) right = mid;
                else left = mid+1;
            }
            else//right side sorted
            {
                if(A[mid] <= target && target <= A[right]) left = mid;
                else right = mid-1;
            }
        }
        if(right >= 0 && right < n && A[right] == target) return right;
        else return -1;
    }
};