zoukankan      html  css  js  c++  java
  • Codility----PermMissingElem

    Task description

    A zero-indexed array A consisting of N different integers is given. The array contains integers in the range [1..(N + 1)], which means that exactly one element is missing.

    Your goal is to find that missing element.

    Write a function:

    class Solution { public int solution(int[] A); }

    that, given a zero-indexed array A, returns the value of the missing element.

    For example, given array A such that:

    A[0] = 2 A[1] = 3 A[2] = 1 A[3] = 5

    the function should return 4, as it is the missing element.

    Assume that:

    • N is an integer within the range [0..100,000];
    • the elements of A are all distinct;
    • each element of array A is an integer within the range [1..(N + 1)].

    Complexity:

    • expected worst-case time complexity is O(N);
    • expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

    Elements of input arrays can be modified.

      

    Solution
     
    Programming language used: Java
    Total time used: 4 minutes
    Code: 10:43:42 UTC, java, final, score:  100
    // you can also use imports, for example:
    // import java.util.*;
    
    // you can write to stdout for debugging purposes, e.g.
    // System.out.println("this is a debug message");
    import java.util.Arrays;
    class Solution {
        public int solution(int[] A) {
            // write your code in Java SE 8
            int res = 0;
            int max = A.length;
            if(max==0)
                return 1;
            Arrays.sort(A);
            for(int i=0; i<max; i++){
                res = (i+1)^A[i];
                if(res != 0)
                    return (i+1);
            }
            return max+1;
        }
    }
  • 相关阅读:
    Django重新构造User模型
    在docker中添加mysql在通过远程机器的访问
    php基础笔记
    mysql基础笔记整理
    redis的配置安装与使用
    c++实现对两个有序链表的连接
    java的网络编程(TCP)
    无心制作
    nacos配置服务
    声明式远程调用OpenFeign(微服务调用微服务)
  • 原文地址:https://www.cnblogs.com/samo/p/6775688.html
Copyright © 2011-2022 走看看