【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.

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

通过区间首末元素的大小关系可以确定在该区间内是否有rotate。若进行过翻转则首元素必大于末元素。

复杂在于边界的确定，要注意什么时候需要包括“=”。