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‎README.adoc

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//|{doc_base_url}/1005-maximize-sum-of-array-after-k-negations.adoc[题解]
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//|Easy
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//|
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//
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//|{counter:codes}
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//|{leetcode_base_url}/clumsy-factorial/[1006. Clumsy Factorial^]
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//|{source_base_url}/_1006_ClumsyFactorial.java[Java]
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//|{doc_base_url}/1006-clumsy-factorial.adoc[题解]
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//|Medium
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//|
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//
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|{counter:codes}
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|{leetcode_base_url}/clumsy-factorial/[1006. Clumsy Factorial^]
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|{source_base_url}/_1006_ClumsyFactorial.java[Java]
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|{doc_base_url}/1006-clumsy-factorial.adoc[题解]
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|Medium
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|
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//|{counter:codes}
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//|{leetcode_base_url}/minimum-domino-rotations-for-equal-row/[1007. Minimum Domino Rotations For Equal Row^]
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//|{source_base_url}/_1007_MinimumDominoRotationsForEqualRow.java[Java]

‎docs/1006-clumsy-factorial.adoc

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[#1006-clumsy-factorial]
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= 1006. Clumsy Factorial
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= 1006. 笨阶乘
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{leetcode}/problems/clumsy-factorial/[LeetCode - Clumsy Factorial^]
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https://leetcode.cn/problems/clumsy-factorial/[LeetCode - 1006. 笨阶乘 ^]
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Normally, the factorial of a positive integer `n` is the product of all positive integers less than or equal to `n`. For example, `factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1`.
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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`。
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We instead make a _clumsy factorial:_ using the integers in decreasing order, we swap out the multiply operations for a fixed rotation of operations: multiply (*), divide (/), add (+) and subtract (-) in this order.
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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(-)。
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For example, `clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1`. However, these operations are still applied using the usual order of operations of arithmetic: we do all multiplication and division steps before any addition or subtraction steps, and multiplication and division steps are processed left to right.
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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。
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Additionally, the division that we use is _floor division_ such that `10 * 9 / 8` equals `11`. This guarantees the result is an integer.
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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。
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`<font face="sans-serif, Arial, Verdana, Trebuchet MS">Implement the clumsy` function as defined above: given an integer `N`, it returns the clumsy factorial of `N`.
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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。
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*Example 1:*
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*示例 1：*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* 4
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*Output:* 7
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*Explanation:* 7 = 4 * 3 / 2 + 1
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----
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....
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输入：4
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输出：7
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解释：7 = 4 * 3 / 2 + 1
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....
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*Example 2:*
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*示例 2：*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* 10
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*Output:* 12
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*Explanation:* 12 = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1
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----
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....
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输入：10
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输出：12
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解释：12 = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1
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....
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*Note:*
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*提示：*
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. `+1 <= N <= 10000+`
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. `-2^31^ \<= answer \<= 2^31^ - 1` （答案保证符合 32 位整数。）
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. `1 <= N <= 10000`
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. `-2^31 <= answer <= 2^31 - 1` (The answer is guaranteed to fit within a 32-bit integer.)
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== 思路分析
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看官方题解，可以把减法改成“加负数”，这样就可以极大地减少判断，可以直接按照“遇到乘除立即算，遇到加减先入栈”处理。
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image::images/1006-10.gif[{image_attr}]
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[[src-1006]]
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[tabs]
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====
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一刷::
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_1006_ClumsyFactorial.java[tag=answer]
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----
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--
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// 二刷::
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// +
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// --
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// [{java_src_attr}]
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// ----
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// include::{sourcedir}/_1006_ClumsyFactorial_2.java[tag=answer]
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// ----
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// --
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====
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== 参考资料
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. https://leetcode.cn/problems/clumsy-factorial/solutions/692629/ben-jie-cheng-by-leetcode-solution-deh2/[1006. 笨阶乘 - 官方题解^] -- 可以把减法改成“加负数”，这样就可以极大地减少判断。
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. https://leetcode.cn/problems/clumsy-factorial/solutions/693117/fu-xue-ming-zhu-yu-dao-cheng-chu-li-ji-s-furg/[1006. 笨阶乘 - 【负雪明烛】遇到乘除立即算，遇到加减先入栈^]

‎docs/images/1006-10.gif

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‎docs/index.adoc

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// include::1005-maximize-sum-of-array-after-k-negations.adoc[leveloffset=+1]
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// include::1006-clumsy-factorial.adoc[leveloffset=+1]
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include::1006-clumsy-factorial.adoc[leveloffset=+1]
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// include::1007-minimum-domino-rotations-for-equal-row.adoc[leveloffset=+1]
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‎logbook/202503.adoc

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|{doc_base_url}/1217-minimum-cost-to-move-chips-to-the-same-position.adoc[题解]
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|❌ 题目理解错误！`position[i]` 中的数字是筹码的位置。题目意思是奇数到奇数和偶数到偶数免费，但是紧挨的两个数字移动收费，所以，统计奇偶数的个数，看哪个小就移动哪个。
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|{counter:codes2503}
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|{leetcode_base_url}/clumsy-factorial/[1006. 笨阶乘^]
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|{doc_base_url}/1006-clumsy-factorial.adoc[题解]
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|✅ 栈。代码写出来了，但是好烂！
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|===
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截止目前，本轮练习一共完成 {codes2503} 道题。

‎src/main/java/com/diguage/algo/leetcode/_1006_ClumsyFactorial.java

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package com.diguage.algo.leetcode;
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import java.util.ArrayDeque;
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import java.util.Deque;
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import java.util.function.BiFunction;
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public class _1006_ClumsyFactorial {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2025-05-23 14:54:02
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*/
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public int clumsy(int n) {
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char[] operators = new char[]{'*', '/', '+', '-'};
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BiFunction<Integer, Integer, Integer>[] ops = new BiFunction[4];
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ops[0] = (x, y) -> x * y;
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ops[1] = (x, y) -> x / y;
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ops[2] = Integer::sum;
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ops[3] = (x, y) -> x - y;
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int idx = 0;
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int result = 0;
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Deque<Integer> stack = new ArrayDeque<>(n);
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stack.offer(n--);
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while (!stack.isEmpty() && n > 0) {
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if (idx == 2 && stack.size() == 2) {
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Integer b = stack.pollLast();
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Integer a = stack.pollLast();
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stack.offer(ops[3].apply(a, b));
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continue;
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}
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if (idx == 0 || idx == 1 || idx == 2) {
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Integer pre = stack.pollLast();
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Integer res = ops[idx].apply(pre, n--);
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stack.offerLast(res);
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} else {
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stack.offerLast(n--);
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}
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idx++;
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idx %= operators.length;
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}
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if (stack.size() == 1) {
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result = stack.pollLast();
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} else {
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Integer b = stack.pollLast();
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Integer a = stack.pollLast();
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result = ops[3].apply(a, b);
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}
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return result;
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}
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// end::answer[]
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public static void main(String[] args) {
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new _1006_ClumsyFactorial().clumsy(4);
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}
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}

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