[LeetCode]Search in Rotated Sorted Array

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.

分析：二分查找，难度在于左右边界的确定。

class Solution {
public:
int search(int A[],int n,int target) {
int low = 0, high = n-1;
while (low != high){
const int mid = low + (high - low) / 2;
if (A[mid] == target)
return mid;
if (A[low] <= A[mid]){
if (A[low]<=target && A[mid]>target){
high = mid;
}
else
low = mid+1;
}
else{
if (target > A[mid] && target <= A[high]){
low = mid+1;
}
else{
high = mid;
}
}
}
return -1;
}
};
class Solution1 {
int binary_search(int A[], int low, int high, int target){
while (low <= high){
int mid = low + (high - low) / 2;
if (A[mid] == target)
return mid;
else if (A[mid] < target)
low = mid + 1;
else
high = mid - 1;
}

return -1;
}

public:
// assume it is an ascending array
int search(int A[], int n, int target) {
int low = 0, high = n - 1;

while (low <= high){
if (A[low] <= A[high]) // It will be satified at the end at least.
return binary_search(A, low, high, target);

int mid = low + (high - low) / 2;
if (A[mid] == target)
return mid;
else if (A[mid] < A[low]){
// [low, mid] is not ascending, [mid, high] is ascending
if (target > A[mid] && target <= A[high])
low = mid + 1;
else
high = mid - 1;
}
else{
// [low, mid] is ascending, [mid, high] is not ascending
if (target >= A[low] && target < A[mid])
high = mid - 1;
else
low = mid + 1;
}
}
return -1;
}
};

int main(){
Solution solution;

{
int a[] = { 4, 5, 6, 7, 0, 1, 2, };
int target = 5;
cout << solution.search(a, sizeof(a) / sizeof(a[0]), target)<<endl;
}
getchar();
return 0;

}