---

comments: true
difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/lcof/%E9%9D%A2%E8%AF%95%E9%A2%9864.%20%E6%B1%821%2B2%2B%E2%80%A6%2Bn/README.md

面试题 64. 求 1+2+…+n

题目描述

求 1+2+...+n ，要求不能使用乘除法、for、while、if、else、switch、case等关键字及条件判断语句（A?B:C）。

 

示例 1：

输入: n = 3
输出: 6

示例 2：

输入: n = 9
输出: 45

 

限制：

• 1 <= n <= 10000

解法

方法一

Python3

class Solution:def sumNums(self, n: int) -> int:# n 是 0，所以 0 and (0 + self.sumNums(-1)) 结果是 0return n and (n + self.sumNums(n - 1)) 

Java

class Solution {public int sumNums(int n) {int s = n;boolean t = n > 0 && (s += sumNums(n - 1)) > 0;return s;}
}

C++

class Solution {
public:int sumNums(int n) {n && (n += sumNums(n - 1));return n;}
};

Go

func sumNums(n int) int {s := 0var sum func(int) boolsum = func(n int) bool {s += nreturn n > 0 && sum(n-1)}sum(n)return s
}

TypeScript

var sumNums = function (n: number): number {return n && n + sumNums(n - 1);
};

Rust

impl Solution {pub fn sum_nums(mut n: i32) -> i32 {n != 0&& ({n += Solution::sum_nums(n - 1);},true,).1;n}
}

JavaScript

/*** @param {number} n* @return {number}*/
var sumNums = function (n) {return (n ** 2 + n) >> 1;
};

C#

public class Solution {public int result;public int SumNums(int n) {helper(n);return result;}public bool helper(int n) {result += n;return n == 0 || helper(n - 1);}
}

Swift

class Solution {func sumNums(_ n: Int) -> Int {var s = nlet _ = n > 0 && { s += sumNums(n - 1); return true }()return s}
}