33. Search in Rotated Sorted Array (M)

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本文介绍了一种在旋转后的有序数组中查找目标值的算法。数组原本按升序排列，但未知位置进行了旋转。通过一次或两次二分查找，可以在O(logn)的时间复杂度内找到目标值或返回-1。文章提供了两种实现方式：一种是先找到两个递增序列的边界，然后在相应的序列中查找；另一种是直接在一个二分查找中完成整个过程。

Search in Rotated Sorted Array (M)

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 = 0
Output: 4

题意

将一递增数列的随机后半部分与前半部分换位，得到新数组，在新数组中查找目标值。

思路

比较简单的做法是先用一个二分找到两个递增序列的边界，再用一个二分在对应的递增序列里查找目标值。

也可以只用一个二分完成操作：求出mid，先考虑mid落在右递增区间的情况，如果target也落在右递增区间且比nums[mid]大，说明只需要继续向右查找，令 left = mid + 1 即可，不然只要向左查找，令 right = mid - 1；同理，对于mid落在左递增区间的情况进行处理。

在这里插入图片描述

代码实现 - 不求边界

class Solution {
    public int search(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;
        
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] == target) {
                return mid;
            }
            // 通过与第一个值相比较来判断mid落在左区间还是右区间
            if (nums[mid] < nums[left]) {
                if (target <= nums[right] && target > nums[mid]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            } else {
                if (target >= nums[left] && target < nums[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            }
        }
        
        return -1;
    }
}

代码实现 - 先求边界

class Solution {
    public int search(int[] nums, int target) {
        if (nums.length == 0) {
            return -1;
        }
        
        int left = 0;
        int right = nums.length - 1;
        int split = nums.length - 1;		// 默认不存在边界
        
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] == target) {
                return mid;
            }
            if (nums[mid] >= nums[left]) {
                if (mid + 1 < nums.length && nums[mid + 1] < nums[mid]) {
                    split = mid;
                    break;
                } else {
                    left = mid + 1;
                }
            } else {
                if (mid - 1 >= 0 && nums[mid - 1] > nums[mid]) {
                    split = mid - 1;
                    break;
                } else {
                    right = mid - 1;
                }
            }
        }
		
        // 选择一个递增区间进行二分查找
        if (target >= nums[0]) {
            return binarySearch(nums, 0, split, target);
        } else {
            return binarySearch(nums, split + 1, nums.length - 1, target);
        }
    }

    private int binarySearch(int[] nums, int left, int right, int target) {
        while (left <= right) {
            int mid = (left + right) / 2;
            if (target > nums[mid]) {
                left = mid + 1;
            } else if (target < nums[mid]) {
                right = mid - 1;
            } else {
                return mid;
            }
        }
        return -1;
    }
}