[LeetCode] 702. Search in a Sorted Array of Unknown Size

This is an interactive problem.

You have a sorted array of unique elements and an unknown size. You do not have an access to the array but you can use the ArrayReader interface to access it. You can call ArrayReader.get(i) that:

• returns the value at the ith index (0-indexed) of the secret array (i.e., secret[i]), or
• returns 231 - 1 if the i is out of the boundary of the array.

You are also given an integer target.

Return the index k of the hidden array where secret[k] == target or return -1 otherwise.

You must write an algorithm with O(log n) runtime complexity.

Example 1:

Input: secret = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in secret and its index is 4.

Example 2:

Input: secret = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in secret so return -1.

Constraints:

• 1 <= secret.length <= 104
• -104 <= secret[i], target <= 104
• secret is sorted in a strictly increasing order.

在未知大小的有序数组中搜索。

题干即是题意，给一个有序数组但是不告知数组长度，请找一个 target 数字。

思路是用二分法。先设一个变量 high，因为不确定 input 数组的长度所以只能每次乘以二去试探 high 位置上的数字和 target 数字的大小关系。

时间O(logn)

空间O(1)

JS实现

/**
 * @param {ArrayReader} reader
 * @param {number} target
 * @return {number}
 */
var search = function (reader, target) {
    let hi = 1;
    while (reader.get(hi) < target) {
        hi = hi << 1;
    }
    let low = hi >> 1;
    while (low <= hi) {
        let mid = Math.floor(low + (hi - low) / 2);
        if (reader.get(mid) > target) {
            hi = mid - 1;
        } else if (reader.get(mid) < target) {
            low = mid + 1;
        } else {
            return mid;
        }
    }
    return -1;
};

Java实现

class Solution {
    public int search(ArrayReader reader, int target) {
        int hi = 1;
        while (reader.get(hi) < target) {
            hi = hi << 1;
        }
        int low = hi >> 1;
        while (low <= hi) {
            int mid = low + (hi - low) / 2;
            if (reader.get(mid) > target) {
                hi = mid - 1;
            } else if (reader.get(mid) < target) {
                low = mid + 1;
            } else {
                return mid;
            }
        }
        return -1;
    }
}

LeetCode 题目总结