Can you solve this real interview question? Reach a Number - You are standing at position 0 on an infinite number line. There is a destination at position target. You can make some number of moves numMoves so that: * On each move, you can either go le
int i=0 ; while(i*(i+1) < 2*target ) { i++; } if(i*(i+1)/2 == target) return i ; else { if((i*(i+1)/2-target)%2 == 0) return i; else { if(i%2==0) return i+1 ; else return i+2 ; } } } }; C/C++基本语法学习 STL C++ primer 分类: leetcode 好文...
https://github.com/grandyang/leetcode/issues/754 参考资料: https://leetcode.com/problems/reach-a-number/ https://leetcode.com/problems/reach-a-number/discuss/112969/C++-O(1)-Solution-without-loop https://leetcode.com/problems/reach-a-number/discuss/112992/Learn-from-other-with-my-explanatio...
LeetCode 754. Reach a Number (Java版; Medium) 题目描述 AI检测代码解析 You are standing at position 0 on an infinite number line. There is a goal at position target. On each move, you can either go left or right. During the n-th move (starting from 1), you take n steps. ...
【Leetcode_easy】754.Reacha Number problem 754.Reacha Number solution1: solution2: solution3:solution1的精简版; 参考 1. Leetcode_easy_754.Reacha Number; 2. Grandyang; 完 其它 原创 mb62c788fd198da 2022-07-09 00:50:04 25阅读
[LeetCode] Reach a Number - Python 转载自:https://blog.csdn.net/guoziqing506/article/details/79873193 题目描述:我简单将题目用汉语描述一下。说初始位置在0,给定一个目标值target(target是一个整数),现在让你每次走一步,每次走的步长和已经走的步数相等(第一次走1,第二次走2,。。。),每次可以向左...
在leetcode刷题中,自己定义的一个新函数时,会遇到这样的错误: error: control reaches end of non-void function [-Werror=return-type] 比如: 这是一个冒泡排序,void没有返回值的,在调用的时候是这样子的: sort(nums, numszise) 结果: 会出现上面那个错误,但只要在函数后面加上return nums就可...编译...
2.abs(target)是奇数,需要满足上面的a和d两个条件; 代码如下: varreachNumber =function(target) {if(target < 0){ target= -target }if(target == 1 || target == -1){return1}varisOdd = target%2varcount = 1;vart = 1;while(count++) { ...
leetcode 754. 到达终点数字(Reach a Number) 目录 题目描述: 解法: 题目描述: 在一根无限长的数轴上,你站在0的位置。终点在target的位置。 每次你可以选择向左或向右移动。第 n 次移动(从 1 开始),可以走 n 步。 返回到达终点需要的最小移动次数。
1+...-(sum-target+1)/2+...k=target-1 再考虑k的奇偶性。 3.1 如果k是偶数并且k>sum-target+1 那么1+...-(sum-target+1)/2+...-(k/2)...+k+(k+1)=target 相当于在1+2...+k+(k+1)减去了sum-target+1和k。 等价于sum+(k+1)-sum+target-1-k===>target也就是答案是k+1. ...