zoukankan      html  css  js  c++  java
  • lintcode447- Search in a Big Sorted Array- medium

    Given a big sorted array with positive integers sorted by ascending order. The array is so big so that you can not get the length of the whole array directly, and you can only access the kth number by ArrayReader.get(k) (or ArrayReader->get(k) for C++). Find the first index of a target number. Your algorithm should be in O(log k), where k is the first index of the target number.

    Return -1, if the number doesn't exist in the array.

     Notice

    If you accessed an inaccessible index (outside of the array), ArrayReader.get will return 2,147,483,647.

    Example

    Given [1, 3, 6, 9, 21, ...], and target = 3, return 1.

    Given [1, 3, 6, 9, 21, ...], and target = 4, return -1.

    Challenge 

    O(log k), k is the first index of the given target number.

    难点在于end不能直接取到,所以一开始二倍增长地去找一个可能性end(target <= reader.get(end))。接下来就是普通的二分查找了。

    public class Solution {
        /*
         * @param reader: An instance of ArrayReader.
         * @param target: An integer
         * @return: An integer which is the first index of target.
         */
        public int searchBigSortedArray(ArrayReader reader, int target) {
    
            final int MAX = 2147483647;
            int start = 0;
            int end = 1;
            int flip = 1;
    
            while (reader.get(end) < target){
                end += flip;
                flip *= 2;
                if (flip == 2 && reader.get(end) == MAX){
                    return -1;
                }
                if (reader.get(end) == MAX){
                    flip /= 2;
                    end -= flip;
                    flip = 1;
                }
            }
    
            while (start + 1 < end){
                int mid = start + (end - start) / 2;
                if (target < reader.get(mid)){
                    end = mid;
                } else if (target == reader.get(mid)){
                    end = mid;
                } else {
                    start = mid;
                }
            }
            if (target == reader.get(start)){
                return start;
            }
            if (target == reader.get(end)){
                return end;
            }
            return -1;
        }
    }
  • 相关阅读:
    CPU,MPU,MCU,SOC,SOPC联系与差别
    call && jmp 指令
    认识OD的两种断点
    VB逆向
    ASProtect.SKE.2.11 stolen code解密
    破解之寻找OEP[手动脱壳](2)
    破解之寻找OEP[手动脱壳](1)
    破解常用断点设置
    VB断点大全
    API断点大全
  • 原文地址:https://www.cnblogs.com/jasminemzy/p/7580057.html
Copyright © 2011-2022 走看看