Leetcode 33. Search in Rotated Sorted Array

题目链接

https://leetcode.com/problems/search-in-rotated-sorted-array/description/

题目描述

Suppose an array sorted in ascending order 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.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

题解

根据题目要求的时间复杂度,这题需要使用二分查找。每次二分查找之后,我们找到该数组中单调的部分来和目标元素相比较。这样可以避免漏掉一些情况。

代码

class Solution {
    public int search(int[] a, int target) {
        if (a == null || a.length == 0) {
            return -1;
        }
        int lo = 0, hi = a.length - 1;
        while (lo <= hi) {
            int mid = (lo + hi) >> 1;
            if (a[mid] == target) { return mid; }
            if (a[mid] > a[hi]) {
                if (a[lo] <= target && target < a[mid]) {
                    hi = mid - 1;
                } else {
                    lo = mid + 1;
                }
            } else {
                if (target > a[mid] && target <= a[hi]) {
                    lo = mid + 1;
                }else {
                    hi = mid - 1;
                }
            }
        }
        return -1;
    }
}

原文地址:https://www.cnblogs.com/xiagnming/p/9598540.html