[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.

 

二分搜索来找到转折点，也就是最小数的位置。对二分搜索要稍作修改，当a[left] <= a[mid]时，可以肯定a[left..mid]是升序的，那么a[left]是最小的，也肯能最小的在a[mid+1..right]中，

所以要比较a[left]和min(a[mid+1..right])，同样对于a[left]>a[mid]，做同样的处理。

class Solution {
public:
    int findPos(int a[], int left, int right)
    {
        if (left > right)
            return -1;
            
        int mid = left + (right - left) / 2;
        
        if (a[left] <= a[mid])
        {
            int pos = findPos(a, mid + 1, right);
            
            if (pos == -1)
                return left;
            else
                return a[left] < a[pos] ? left : pos; 
        }
        else
        {
            int pos = findPos(a, left, mid - 1);
            
            if (pos == -1)
                return mid;
            else
                return a[pos] < a[mid] ? pos : mid;
        }
    }
    
    int bsearch(int a[], int left, int right, int key)
    {
        if (left > right)
            return -1;
            
        int mid = left + (right - left) / 2;
        
        if (a[mid] == key)
            return mid;
        else if (a[mid] < key)
            return bsearch(a, mid + 1, right, key);
        else
            return bsearch(a, left, mid - 1, key);
    }
    
    int search(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int pos = findPos(A, 0, n - 1);
        
        int index = bsearch(A, 0, pos - 1, target);
        if (index != -1)
            return index;
            
        return bsearch(A, pos, n - 1, target);
    }
};