思路：一共4个数字，共需要3个运算符，可以构造一个二叉树，没有子节点的节点的为值，有叶子节点的为运算符
例如数字{1, 2, 3, 4}，其中一种解的二叉树形式如下所示：
因此可以遍历所有二叉树可能的形式，4个数的全排列，从4种运算符中挑选3种运算符（运算符可以重复）
核心步骤1：需要遍历所有二叉树的可能，参考Eric Lippert的方法
class Node
{
public Node Left { get; private set; }
public Node Right { get; private set; }
public Node(Node left, Node right)
{
this.Left = left;
this.Right = right;
}
} static IEnumerable<Node> AllBinaryTrees(int size)
{
if (size == )
return new Node[] { null };
return from i in Enumerable.Range(, size)
from left in AllBinaryTrees(i)
from right in AllBinaryTrees(size - - i)
select new Node(left, right);
}
核心步骤2：对于任意一个二叉树，构造表达式树
static Expression Build(Node node, List<double> numbers, List<Func<Expression, Expression, BinaryExpression>> operators)
{
var iNum = ;
var iOprt = ; Func<Node, Expression> f = null;
f = n =>
{
Expression exp;
if (n == null)
exp = Expression.Constant(numbers[iNum++]);
else
{
var left = f(n.Left);
var right = f(n.Right);
exp = operators[iOprt++](left, right);
}
return exp;
};
return f(node);
}
核心步骤3：遍历4个数字的全排列，全排列参考这里
static IEnumerable<List<T>> FullPermute<T>(List<T> elements)
{
if (elements.Count == )
return EnumerableOfOneElement(elements); IEnumerable<List<T>> result = null;
foreach (T first in elements)
{
List<T> remaining = elements.ToArray().ToList();
remaining.Remove(first);
IEnumerable<List<T>> fullPermuteOfRemaining = FullPermute(remaining); foreach (List<T> permute in fullPermuteOfRemaining)
{
var arr = new List<T> { first };
arr.AddRange(permute); var seq = EnumerableOfOneElement(arr);
if (result == null)
result = seq;
else
result = result.Union(seq);
}
}
return result;
} static IEnumerable<T> EnumerableOfOneElement<T>(T element)
{
yield return element;
}
例如有四个数字{1, 2, 3, 4}，它的全排列如下：
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
, , ,
核心步骤4：从4种运算符中挑选3个运算符
static IEnumerable<IEnumerable<Func<Expression, Expression, BinaryExpression>>> OperatorPermute(List<Func<Expression, Expression, BinaryExpression>> operators)
{
return from operator1 in operators
from operator2 in operators
from operator3 in operators
select new[] { operator1, operator2, operator3 };
}
最后是Main函数：
static void Main(string[] args)
{
List<double> numbers = new List<double> { , , , };
var operators = new List<Func<Expression, Expression, BinaryExpression>> {
Expression.Add,Expression.Subtract,Expression.Multiply,Expression.Divide
}; foreach (var operatorCombination in OperatorPermute(operators))
{
foreach (Node node in AllBinaryTrees())
{
foreach (List<double> permuteOfNumbers in FullPermute(numbers))
{
Expression expression = Build(node, permuteOfNumbers, operatorCombination.ToList());
Func<double> compiled = Expression.Lambda<Func<double>>(expression).Compile();
try
{
var value = compiled();
if (Math.Abs(value - ) < 0.01)
Console.WriteLine("{0} = {1}", expression, value);
}
catch (DivideByZeroException) { }
}
}
}
Console.Read();
}
计算结果：
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( * ( + ( + ))) =
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( * (( + ) + )) =
( * (( + ) + )) =
( * (( + ) + )) =
( * (( + ) + )) =
( * (( + ) + )) =
(( + ) * ( + )) =
(( + ) * ( + )) =
(( + ) * ( + )) =
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(( + ) * ( + )) =
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(( + ( + )) * ) =
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((( + ) + ) * ) =
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对于一些平时口算相对稍难的一些组合也是毫无压力，例如{1, 5, 5, 5}， {3, 3, 7, 7}， {3, 3, 8, 8}，有兴趣的看官口算试试 :)