Lintcode062.Search in Rotated Sorted Array

Posted 2020-09-17 Vincent丶

题目：

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.

Example

For [4, 5, 1, 2, 3] and target=1, return 2.

For [4, 5, 1, 2, 3] and target=0, return -1.

题解：

class Solution {
public:
    /**
     * @param nums: a rotated sorted array
     * @return: the minimum number in the array
     */
    int search(vector<int> &A, int target) {
        if (A.empty()) {
            return -1;
        }
        int start = 0, end = A.size() - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (A[mid] == target) {
                return mid;
            } 
            if (A[start] < A[mid]) {
                if (A[start] <= target && target <= A[mid]) {
                    end = mid;
                } else {
                    start = mid;
                }
            } else {
                if (A[mid] <= target && target <= A[end]) {
                    start = mid;
                } else {
                    end = mid;
                }
            }
        }
        
        if (A[start] == target) {
            return start;
        } else if (A[end] == target){
            return end;
        }
        
        return -1;
    }
};