LeetCode 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.

思路：首先对于left和right看看，A[left] <= A[right]的话那么证明这段数组是有序的，否则的话，就是一个有折点的数组，再在左右两段中继续二分

class Solution {
public:
    int search(int A[], int n, int target) {
       return find(A, 0, n-1, target); 
    }

    int find(int A[], int l, int r, int target) {
        if (l > r) return -1;

        int ans = -1;
        if (A[l] <= A[r]) {
            int left = l, right = r;
            while (left <= right) {
                int mid = left + right >> 1;
                if (A[mid] == target) {
                    ans = mid;
                    break;
                }
                if (A[mid] > target) 
                    right = mid - 1;
                else left = mid + 1;
            }
        } else {
            int mid = l + r >> 1;
            if (A[mid] == target) ans = mid;
            else {
                ans = find(A, l, mid - 1, target);
                ans = ans == -1 ? find(A, mid + 1, r, target) : ans;
            }
        }

        return ans;
    }
};

LeetCode Search in Rotated Sorted Array

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原文地址：http://blog.csdn.net/u011345136/article/details/43991047