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通常,正整数n的阶乘是所有小于或等于 n 的正整数的乘积。例如,factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1。
n
factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
相反,我们设计了一个笨阶乘clumsy:在整数的递减序列中,我们以一个固定顺序的操作符序列来依次替换原有的乘法操作符:乘法(*),除法(/),加法(+)和减法(-)。
clumsy:
例如,clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1。然而,这些运算仍然使用通常的算术运算顺序:我们在任何加、减步骤之前执行所有的乘法和除法步骤,并且按从左到右处理乘法和除法步骤。
clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1
另外,我们使用的除法是地板除法(floor division),所以 10 * 9 / 8 等于11。这保证结果是一个整数。
(floor division)
10 * 9 / 8
11
实现上面定义的笨函数:给定一个整数 N,它返回 N 的笨阶乘。
输入:4 输出:7 解释:7 = 4 * 3 / 2 + 1
输入:10 输出:12 解释:12 = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/clumsy-factorial 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
The text was updated successfully, but these errors were encountered:
怎么复杂怎么来,哈哈哈
func clumsy(N int) int { var stack []string operator := [4]string{"*", "/", "+", "-"} index := 0 for N > 0 { stack = append(stack, strconv.Itoa(N)) if N == 1 { break } stack = append(stack, operator[index]) N-- index++ if index == 4 { index = 0 } } return calculate(stack) } func calculate(exp []string) int { var stack []string for _, v := range exp { if len(stack) == 0 { stack = append(stack, v) continue } express := stack[len(stack)-1] if express == "*" || express == "/" { pre, _ := strconv.Atoi(stack[len(stack)-2]) cur, _ := strconv.Atoi(v) stack = stack[:len(stack)-2] if express == "*" { stack = append(stack, strconv.Itoa(pre*cur)) } else { stack = append(stack, strconv.Itoa(pre/cur)) } } else { stack = append(stack, v) } } ret,_ := strconv.Atoi(stack[0]) for i := 1; i < len(stack); i++ { if stack[i] == "+" { cur,_ := strconv.Atoi(string(stack[i+1])) ret += cur } else { cur,_ := strconv.Atoi(string(stack[i+1])) ret -= cur } i++ } return ret }
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通常,正整数
n
的阶乘是所有小于或等于 n 的正整数的乘积。例如,factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
。相反,我们设计了一个笨阶乘
clumsy:
在整数的递减序列中,我们以一个固定顺序的操作符序列来依次替换原有的乘法操作符:乘法(*),除法(/),加法(+)和减法(-)。例如,
clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1
。然而,这些运算仍然使用通常的算术运算顺序:我们在任何加、减步骤之前执行所有的乘法和除法步骤,并且按从左到右处理乘法和除法步骤。另外,我们使用的除法是地板除法
(floor division)
,所以10 * 9 / 8
等于11
。这保证结果是一个整数。实现上面定义的笨函数:给定一个整数 N,它返回 N 的笨阶乘。
示例 1:
示例 2:
提示:
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/clumsy-factorial
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
The text was updated successfully, but these errors were encountered: