LeetCode每日一题：search in rotated s

问题描述

Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e.,0 1 2 4 5 6 7might become4 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.

问题分析

这道题暴力求解也是可以的，但是正确的做法是找出分界点，对左右两边进行二分查找，这样的速度会比较快

代码实现

public int search(int[] A, int target) {
        int position = 0;
        for (int i = 1; i < A.length; i++) {
            if (A[i] < A[i - 1]) {
                position = i - 1;
            }
        }
        int low = 0;
        int high = position;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (A[mid] == target) return mid;
            else if (A[mid] > target) high = mid - 1;
            else if (A[mid] < target) low = mid + 1;
        }
        low = position + 1;
        high = A.length - 1;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (A[mid] == target) return mid;
            else if (A[mid] > target) high = mid - 1;
            else if (A[mid] < target) low = mid + 1;
        }
        return -1;
    }