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【算法学习】剑指 Offer 64. 求1+2+…+n(java / c / c++ / python / go / rust)

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剑指 Offer 64. 求1+2+…+n:

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)。

样例 1

输入: 
	n = 3
输出: 
	6
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样例 2

输入: 
	n = 9
输出: 
	45
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限制

  • 1 <= n <= 10000

分析

  • 常规做法就是乘法,循环,递归。
  • 题目不让用乘法和循环,那就用递归替代;递归需要有终止条件,题目不然用条件判断,我们可以用短路逻辑运算符替代。
  • 递归太慢了,我们可以用等差数列求和公式,就成了计算n * (n + 1) / 2。
  • 除以2可以用向右移位替代。
  • 乘法,我们可以考虑小学时候算乘法的的步骤,a × b,我们就是用b的每一位数去乘以a × 10^i^,i表示b的第几位数,再把结果加起来。那如果b是2进制的每一位,就变成看b的每一位是0还是1,如果是1就加上1 × a × 2^i^,用移位替代就变成a << i。
  • 限制里n最大是10000,2^14^就够了。

题解

java

class Solution {
    public int sumNums(int n) {
        // 2的14次正好超过10000
		// 等差数列求和公式n * (n + 1) / 2
		int ans = 0, a = n, b = n + 1;
		// 没什么卵用,为了让表达式变成语句
		boolean flag;

		flag = ((b & 1) > 0) && (ans += a) > 0;
		a <<= 1;
		b >>= 1;

		flag = ((b & 1) > 0) && (ans += a) > 0;
		a <<= 1;
		b >>= 1;

		flag = ((b & 1) > 0) && (ans += a) > 0;
		a <<= 1;
		b >>= 1;

		flag = ((b & 1) > 0) && (ans += a) > 0;
		a <<= 1;
		b >>= 1;

		flag = ((b & 1) > 0) && (ans += a) > 0;
		a <<= 1;
		b >>= 1;

		flag = ((b & 1) > 0) && (ans += a) > 0;
		a <<= 1;
		b >>= 1;

		flag = ((b & 1) > 0) && (ans += a) > 0;
		a <<= 1;
		b >>= 1;

		flag = ((b & 1) > 0) && (ans += a) > 0;
		a <<= 1;
		b >>= 1;

		flag = ((b & 1) > 0) && (ans += a) > 0;
		a <<= 1;
		b >>= 1;

		flag = ((b & 1) > 0) && (ans += a) > 0;
		a <<= 1;
		b >>= 1;

		flag = ((b & 1) > 0) && (ans += a) > 0;
		a <<= 1;
		b >>= 1;

		flag = ((b & 1) > 0) && (ans += a) > 0;
		a <<= 1;
		b >>= 1;

		flag = ((b & 1) > 0) && (ans += a) > 0;
		a <<= 1;
		b >>= 1;

		flag = ((b & 1) > 0) && (ans += a) > 0;

		return ans >> 1;
    }
}
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c

int sumNums(int n){
    // 2的14次正好超过10000
    // 等差数列求和公式n * (n + 1) / 2
    int ans = 0, a = n, b = n + 1;

    (b & 1) && (ans += a);
    a <<= 1;
    b >>= 1;

    (b & 1) && (ans += a);
    a <<= 1;
    b >>= 1;

    (b & 1) && (ans += a);
    a <<= 1;
    b >>= 1;

    (b & 1) && (ans += a);
    a <<= 1;
    b >>= 1;

    (b & 1) && (ans += a);
    a <<= 1;
    b >>= 1;

    (b & 1) && (ans += a);
    a <<= 1;
    b >>= 1;

    (b & 1) && (ans += a);
    a <<= 1;
    b >>= 1;

    (b & 1) && (ans += a);
    a <<= 1;
    b >>= 1;

    (b & 1) && (ans += a);
    a <<= 1;
    b >>= 1;

    (b & 1) && (ans += a);
    a <<= 1;
    b >>= 1;

    (b & 1) && (ans += a);
    a <<= 1;
    b >>= 1;

    (b & 1) && (ans += a);
    a <<= 1;
    b >>= 1;

    (b & 1) && (ans += a);
    a <<= 1;
    b >>= 1;

    (b & 1) && (ans += a);

    return ans >> 1;
}
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c++

class Solution {
public:
    int sumNums(int n) {
        // 2的14次正好超过10000
        // 等差数列求和公式n * (n + 1) / 2
        int ans = 0, a = n, b = n + 1;

        (b & 1) && (ans += a);
        a <<= 1;
        b >>= 1;

        (b & 1) && (ans += a);
        a <<= 1;
        b >>= 1;

        (b & 1) && (ans += a);
        a <<= 1;
        b >>= 1;

        (b & 1) && (ans += a);
        a <<= 1;
        b >>= 1;

        (b & 1) && (ans += a);
        a <<= 1;
        b >>= 1;

        (b & 1) && (ans += a);
        a <<= 1;
        b >>= 1;

        (b & 1) && (ans += a);
        a <<= 1;
        b >>= 1;

        (b & 1) && (ans += a);
        a <<= 1;
        b >>= 1;

        (b & 1) && (ans += a);
        a <<= 1;
        b >>= 1;

        (b & 1) && (ans += a);
        a <<= 1;
        b >>= 1;

        (b & 1) && (ans += a);
        a <<= 1;
        b >>= 1;

        (b & 1) && (ans += a);
        a <<= 1;
        b >>= 1;

        (b & 1) && (ans += a);
        a <<= 1;
        b >>= 1;

        (b & 1) && (ans += a);

        return ans >> 1;
    }
};
复制代码

python

class Solution:
    def sumNums(self, n: int) -> int:
        # 2的14次正好超过10000
        # 等差数列求和公式n * (n + 1) / 2
        ans = 0
        a = n
        b = n + 1

        (b & 1) and (ans := ans + a)
        a <<= 1
        b >>= 1

        (b & 1) and (ans := ans + a)
        a <<= 1
        b >>= 1

        (b & 1) and (ans := ans + a)
        a <<= 1
        b >>= 1

        (b & 1) and (ans := ans + a)
        a <<= 1
        b >>= 1

        (b & 1) and (ans := ans + a)
        a <<= 1
        b >>= 1

        (b & 1) and (ans := ans + a)
        a <<= 1
        b >>= 1

        (b & 1) and (ans := ans + a)
        a <<= 1
        b >>= 1

        (b & 1) and (ans := ans + a)
        a <<= 1
        b >>= 1

        (b & 1) and (ans := ans + a)
        a <<= 1
        b >>= 1

        (b & 1) and (ans := ans + a)
        a <<= 1
        b >>= 1

        (b & 1) and (ans := ans + a)
        a <<= 1
        b >>= 1

        (b & 1) and (ans := ans + a)
        a <<= 1
        b >>= 1

        (b & 1) and (ans := ans + a)
        a <<= 1
        b >>= 1

        (b & 1) and (ans := ans + a)

        return ans >> 1
复制代码

go

func sumNums(n int) int {
    // 2的14次正好超过10000
	// 等差数列求和公式n * (n + 1) / 2
	ans, a, b := 0, n, n + 1
	addGreatZero := func() bool {
		ans += a
		return ans > 0
	}

	_ = ((b & 1) > 0) && addGreatZero()
	a <<= 1
	b >>= 1

	_ = ((b & 1) > 0) && addGreatZero()
	a <<= 1
	b >>= 1

	_ = ((b & 1) > 0) && addGreatZero()
	a <<= 1
	b >>= 1

	_ = ((b & 1) > 0) && addGreatZero()
	a <<= 1
	b >>= 1

	_ = ((b & 1) > 0) && addGreatZero()
	a <<= 1
	b >>= 1

	_ = ((b & 1) > 0) && addGreatZero()
	a <<= 1
	b >>= 1

	_ = ((b & 1) > 0) && addGreatZero()
	a <<= 1
	b >>= 1

	_ = ((b & 1) > 0) && addGreatZero()
	a <<= 1
	b >>= 1

	_ = ((b & 1) > 0) && addGreatZero()
	a <<= 1
	b >>= 1

	_ = ((b & 1) > 0) && addGreatZero()
	a <<= 1
	b >>= 1

	_ = ((b & 1) > 0) && addGreatZero()
	a <<= 1
	b >>= 1

	_ = ((b & 1) > 0) && addGreatZero()
	a <<= 1
	b >>= 1

	_ = ((b & 1) > 0) && addGreatZero()
	a <<= 1
	b >>= 1

	_ = ((b & 1) > 0) && addGreatZero()

	return ans >> 1
}
复制代码

rust

impl Solution {
    pub fn sum_nums(n: i32) -> i32 {
        let mut ans = 0;
        let mut a = n;
        let mut b = n + 1;

        ((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
        a <<= 1;
        b >>= 1;

        ((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
        a <<= 1;
        b >>= 1;

        ((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
        a <<= 1;
        b >>= 1;

        ((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
        a <<= 1;
        b >>= 1;

        ((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
        a <<= 1;
        b >>= 1;

        ((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
        a <<= 1;
        b >>= 1;

        ((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
        a <<= 1;
        b >>= 1;

        ((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
        a <<= 1;
        b >>= 1;

        ((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
        a <<= 1;
        b >>= 1;

        ((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
        a <<= 1;
        b >>= 1;

        ((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
        a <<= 1;
        b >>= 1;

        ((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
        a <<= 1;
        b >>= 1;

        ((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
        a <<= 1;
        b >>= 1;
        
        ((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);

        ans >> 1
    }

    fn add_great_zero(ans: &mut i32, n: i32) -> bool {
        *ans += n;
        *ans > 0
    }
}
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在这里插入图片描述


原题传送门:https://leetcode-cn.com/problems/qiu-12n-lcof/


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