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
样例 2
输入:
n = 9
输出:
45
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 × 10i,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 × 2i,用移位替代就变成a << i。
- 限制里n最大是10000,214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;
}
}
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;
}
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;
}
};
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
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
}
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
}
}
原题传送门:https://leetcode-cn.com/problems/qiu-12n-lcof/
非常感谢你阅读本文~
放弃不难,但坚持一定很酷~
希望我们大家都能每天进步一点点~
本文由 二当家的白帽子:https://developer.aliyun.com/profile/sqd6avc7qgj7y 博客原创~