Search in Rotated Sorted Array

Description

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

Tags: Arrays, Binary Search

思路

题意是让你从一个旋转过后的递增序列中寻找给定值，找到返回索引，找不到返回-1，我们在下面这组数据中寻找规律。

1 2 4 5 6 7 0
2 4 5 6 7 0 1
4 5 6 7 0 1 2
5 6 7 0 1 2 4
6 7 0 1 2 4 5
7 0 1 2 4 5 6

由于是旋转一次，所以肯定有一半及以上的序列仍然是具有递增性质的，我们观察到如果中间的数比左面的数大的话，那么左半部分序列是递增的，否则右半部分就是递增的，那么我们就可以确定给定值是否在递增序列中，从而决定取舍哪半边。

class Solution {
    public int search(int[] nums, int target) {
        int l = 0, r = nums.length - 1, mid;
        while (l <= r) {
            mid = l + r >>> 1;
            if (nums[mid] == target) return mid;
            else if (nums[mid] >= nums[l]) {
                if (nums[l] <= target && target < nums[mid]) r = mid - 1;
                else l = mid + 1;
            } else {
                if (nums[mid] < target && target <= nums[r]) l = mid + 1;
                else r = mid - 1;
            }
        }
        return -1;
    }
}

结语

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