[Leetcode 88] 33 Search in Rotated Sorted Array

Problem:

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.

Analysis:

The problem is that the monotic property is broken after shiftting the array. Example is as follows:

Original Array: 1 2 3 4 5 6 7 8 9 10

Shifted Array1: 7 8 9 10 1 2 3 4 5 6

Shifted Array2: 5 6 7 8 9 10 1 2 3 4

For shifted array 1, the mid is 1 and for shifted array 2, the mid is 9. If now we search 10 in the two arrays, Which side should we go ? Both of 1 and 9 are less than 10, so it seems we need to go the left part of the array. But this is not ture for array2. And we also can decied which part to go with the addition information of A[0] and A[e].

One way I can think of is if we encount a sorted part of the array, then we use binary search. But if the part of the array is shifted, then we can only search both sides of the array.

Code:

Naive version:

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 s = 0, e = n-1;

while (s <= e) {
int m = (s + e) / 2;

if (A[m] == target)
return m;
else if (A[m] < target) {
if (A[m] <= A[e]) {
if (A[e] < target)
e = m-1;
else
s = m+1;
} else {
s = m+1;
}
} else { // A[m] > target
if (A[m] >= A[s]) {
if (A[s] > target)
s = m + 1;
else
e = m - 1;
} else {
e = m-1;
}

}

}

return -1;
}
};

View Code