Missing number = A – B Let’s write the solution in code. importjava.util.Arrays;publicclassFindMissingNumberFromSeries{publicstaticvoidmain(String[]args){int[]numbers={1,2,3,4,5,6,7,8,9,11,12};intN=numbers[numbers.length-1];//The last element in the arrayintexpectedSum=(N*(N+...
百度试题 结果1 题目Find the missing number: ( ). A: B: C: D: 相关知识点: 试题来源: 解析 C . ( ). ... . 故选.反馈 收藏
Given an array containsNnumbers of 0 ..N, find which number doesn't exist in the array. Example GivenN=3and the array[0, 1, 3], return2. Challenge Do it in-place with O(1) extra memory and O(n) time. 这道题是LeetCode上的原题,请参见我之前的博客Missing Number 丢失的数字。那...
Find the Missing Number 方法一: 数学方法,先找到最大的值,需要比较最大的值和array size, 要是比array size小, 说明最大值missing。 然后用等差数列公式求得如果不缺失值的和,然后再减去array里数的和。 classSolution {public:/** * @param nums: a vector of integers * @return: an integer*/intfin...
Find the missing number. A. 15 B. 9 C. 27 D. 29 相关知识点: 试题来源: 解析 C 正确答案:C 解析:第一个图中下角1的得出是1的立方为l,第二个图中下角8的得出是2的立方为8,那么第三个图中3的立方就应该为27,所以C选项正确。反馈 收藏 ...
Find the missing number in the sequence:3,6,11,20,37,___,135.Answer:___. 答案 因为6=3×2-0;11=6×2-1;20=11×2-2;37=20×2-3;所以下个数应该为:37×2-4=70;验证:70×2-5=135正好是最后一个数;故答案为:70 结果三 题目 Find the missing number in the sequence:3,6,11,20,...
// C program to find the missing number in the array#include <stdio.h>intmain() {intarr[]={1,2,3,5,6};intsize=0;inti=0;intmissing=0; size=sizeof(arr)/sizeof(arr[0]); missing=(size+1)*(size+2)/2;for(i=0; i<size; i++) missing=missing-arr[i]; printf("Missing numbe...
Given an array containsNnumbers of 0 ..N, find which number doesn't exist in the array. Example GivenN=3and the array[0, 1, 3], return2. Challenge Do it in-place with O(1) extra memory and O(n) time. 这道题是LeetCode上的原题,请参见我之前的博客Missing Number 丢失的数字。那...
结果一 题目 【题目】Find the missing number in the pattern.根据规律,找出横线上的数字100,94,88,82,70,64 答案 【解析】76相关推荐 1【题目】Find the missing number in the pattern.根据规律,找出横线上的数字100,94,88,82,70,64 反馈 收藏 ...
counting sort 的思想。理由原array作为hashtable来实现我们需要实现的东西 My code: // suppose 1-n in array with length n, one number missing, one number repeat oncepublicintfindMissingNumber(int[]nums){if(nums==null||nums.length<=1){return-1;}inti=0;intrepeatIndex=-1;while(i<nums.length...