leetcode: Search in Rotated Sorted Array

最新推荐文章于 2021-04-09 16:46:58 发布

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本文探讨了在已排序数组旋转后的场景下，如何应用二分查找算法进行目标值搜索。通过对比和调整查找区间，确保算法正确处理旋转部分的影响，最终返回目标值的索引或指示未找到。

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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.

这道题和一般的二分法查找有些不一样，在二分的时候，不能直接比较 A[mid]和target，而应该比较A[mid]和A[start]或A[end]的大小差别

class Solution {
public:


    // 4 5 6 7 8 1 2 3
    int search(int A[], int n, int target) {
        
        int start = 0;
        int end   = n-1;
        
        while (start <= end)
        {
            int mid = (start+end)/2;
            
            /* 
            * the key point solution of this problem is we can not just compare the mid and target
            * instead we should compare the mid and the start and end
            * the following is wrong answer
            */
            
            /*
            if (A[mid] < A[end])  // left side
            {
                if (A[start] > target)    // pivot is on the 45678123  find 1 
                {
                    start = mid+1;
                }
                else
                {
                    end = mid-1;
                }
            }
            else if (A[mid] < target) // right side
            {
                if (A[mid] < A[end])    //  45678123   find 8  || 513 find 5
                {
                    start = mid+1;
                }
                else
                {
                    start = mid+1;
                }
            }
            else
            {
                return mid;
            }
            */
            
            if (A[mid] == target)
                return mid;
            
            
            if (A[mid] < A[end])  //be caureful, here we need to compare mid and end(start) not target
            {
                if (target > A[mid] && target <= A[end])
                {
                    start = mid+1;
                }
                else
                {
                    end = mid-1;
                }
            }
            else
            {
                if (target >= A[start] && target < A[mid])
                {
                    end = mid-1;
                }
                else
                {
                    start = mid+1;
                }
            }
        }
        
        return -1;
    }
};