Search in Rotated Sorted Array

Search in Rotated Sorted Array

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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.

给出排好序的数组，在某个pivot进行旋转，这样数组变成两段有序，并且第二段的最大值小于第一段的最小值。要求log(n)内找出target的位置。

思路1：找出第一段有序的末尾位置，然后判断在哪一段进行二分查找。

class Solution {
public:
    int search(int A[], int n, int target) {
         int pivot = findPivot(A, n, target);
         if (-1 == pivot) {
             return -1;
         }
         if (target >= A[0] && target <= A[pivot]) {
             return binarySearch(A, 0, pivot, target);
         }
         if (pivot < n - 1) {
             return binarySearch(A, pivot + 1, n - 1, target);
         }
         return -1;
    }
private:
    int findPivot(int A[], int n, int target) {
        int start = 0, end = n - 1;
        if (A[start] <= A[end]) {
            return end;
        }
        while (start <= end) {
            int middle = (start + end) / 2;
            if (middle + 1 <= n && A[middle] > A[middle + 1]) {
                return middle;
            }
            if (A[middle] >= A[start]) {
                start = middle + 1;
            } else {
                end = middle - 1;
            }
        }
        return -1;
    }
    int binarySearch(int A[], int start, int end, int target) {
        while (start <= end) {
            int middle = (start + end) / 2;
            if (target == A[middle]) {
                return middle;
            }
            else if (target > A[middle]) {
                start = middle + 1;
            } else {
                end = middle - 1;
            }
        }
        return -1; // not found
    }
};

这个思路是可以，只是有很多临界条件要考虑。

思路2：直接二分查找，要判断清楚在左边还是右边查找。

class Solution {
public:
    int search(int A[], int n, int target) {
        int start = 0, end = n - 1;
        while (start <= end) {
            int middle = (start + end) / 2;
            
            if (target == A[middle]) {
                return middle;
            }
            
            if (A[middle] < A[end]) {
                if (target > A[middle] && target <= A[end]) {
                    start = middle + 1;
                } else {
                    end = middle - 1;
                }
            } else {
                if (target >= A[start] && target < A[middle]) {
                    end = middle - 1;
                } else {
                    start = middle + 1;
                }
            }
        }
        return -1;
    }
};