给定一个只包含0到9的数字和一个整数值的字符串, 目标 .找出可以计算的表达式的数量 目标 在给定的数字串中使用二进制运算符+、–和*。
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Input : "123", Target : 6Output : {“1+2+3”, “1*2*3”}Input : “125”, Target : 7Output : {“1*2+5”, “12-5”}
这个问题可以通过将所有可能的二进制运算符放在数字中间并计算它们,然后检查它们是否计算到目标值来解决。
- 在编写递归代码时,我们需要将这些变量作为递归方法的参数——结果向量、输入字符串、当前表达式字符串、目标值、处理输入的位置、当前评估值和评估中的最后一个值。
- 由于乘法运算,最后一个值保留在递归中,在进行乘法运算时,我们需要最后一个值来正确计算。
请参见下面的示例,以便更好地理解——
Input is 125, suppose we have reached till 1+2 now,Input = “125”, current expression = “1+2”, position = 2, current val = 3, last = 2Now when we go for multiplication, we need last value for evaluation as follows:current val = current val - last + last * current valFirst we subtract last and then add last * current val for evaluation, new last is last * current val.current val = 3 – 2 + 2*5 = 11last = 2*5 = 10
在下面的代码中需要注意的另一件事是,我们忽略了所有从0开始的数字,将一个条件作为循环中的第一个条件,这样我们就不会处理03、05等数字。 请参阅使用CXString()函数,该函数将C++字符串转换为C char数组,该函数用于下面的代码中,因为ATOI()函数希望字符数组作为参数而不是字符串。它将字符数组转换为数字。
C++
// C++ program to find all possible expression which // evaluate to target #include <bits/stdc++.h> using namespace std; // Utility recursive method to generate all possible // expressions void getExprUtil(vector<string>& res, string curExp, string input, int target, int pos, int curVal, int last) { // true if whole input is processed with some // operators if (pos == input.length()) { // if current value is equal to target //then only add to final solution // if question is : all possible o/p then just //push_back without condition if (curVal == target) res.push_back(curExp); return ; } // loop to put operator at all positions for ( int i = pos; i < input.length(); i++) { // ignoring case which start with 0 as they // are useless for evaluation if (i != pos && input[pos] == '0' ) break ; // take part of input from pos to i string part = input.substr(pos, i + 1 - pos); // take numeric value of part int cur = atoi (part.c_str()); // if pos is 0 then just send numeric value // for next recursion if (pos == 0) getExprUtil(res, curExp + part, input, target, i + 1, cur, cur); // try all given binary operator for evaluation else { getExprUtil(res, curExp + "+" + part, input, target, i + 1, curVal + cur, cur); getExprUtil(res, curExp + "-" + part, input, target, i + 1, curVal - cur, -cur); getExprUtil(res, curExp + "*" + part, input, target, i + 1, curVal - last + last * cur, last * cur); } } } // Below method returns all possible expression // evaluating to target vector<string> getExprs(string input, int target) { vector<string> res; getExprUtil(res, "" , input, target, 0, 0, 0); return res; } // method to print result void printResult(vector<string> res) { for ( int i = 0; i < res.size(); i++) cout << res[i] << " " ; cout << endl; } // Driver code to test above methods int main() { string input = "123" ; int target = 6; vector<string> res = getExprs(input, target); printResult(res); input = "125" ; target = 7; res = getExprs(input, target); printResult(res); return 0; } |
Python3
# Python3 program to find all possible expression which # evaluate to target # Utility recursive method to generate all possible # expressions def getExprUtil(res, curExp, _input, target, pos, curVal, last): # true if whole input is processed with some # operators if (pos = = len (_input)): # if current value is equal to target #then only add to final solution # if question is : all possible o/p then just #push_back without condition if (curVal = = target): res.append(curExp) return # loop to put operator at all positions for i in range (pos, len (_input)): # ignoring case which start with 0 as they # are useless for evaluation if (i ! = pos and _input[pos] = = '0' ): break # take part of input from pos to i part = _input[pos: i + 1 ].strip() # take numeric value of part cur = int (part) # if pos is 0 then just send numeric value # for next recursion if (pos = = 0 ): getExprUtil(res, curExp + part, _input, target, i + 1 , cur, cur) # try all given binary operator for evaluation else : getExprUtil(res, curExp + "+" + part, _input, target, i + 1 , curVal + cur, cur) getExprUtil(res, curExp + "-" + part, _input, target, i + 1 , curVal - cur, - cur) getExprUtil(res, curExp + "*" + part, _input, target, i + 1 , curVal - last + last * cur, last * cur) # Below method returns all possible expression # evaluating to target def getExprs(_input, target): res = [] getExprUtil(res, "", _input, target, 0 , 0 , 0 ) return res # method to print result def printResult(res): for i in range ( len (res)): print (res[i],end = " " ) print () # Driver code to test above methods if __name__ = = '__main__' : _input = "123" target = 6 res = getExprs(_input, target) printResult(res) _input = "125" target = 7 res = getExprs(_input, target) printResult(res) |
C#
// C# program to find all possible expression which // evaluate to target using System; using System.Collections.Generic; class GFG { // Utility recursive method to generate all possible expressions static void getExprUtil(List< string > res, string curExp, string input, int target, int pos, int curVal, int last) { // true if whole input is processed with some // operators if (pos == input.Length) { // if current value is equal to target //then only add to final solution // if question is : all possible o/p then just //push_back without condition if (curVal == target) res.Add(curExp); return ; } // loop to put operator at all positions for ( int i = pos; i < input.Length; i++) { // ignoring case which start with 0 as they // are useless for evaluation if (i != pos && input[pos] == '0' ) break ; // take part of input from pos to i string part = input.Substring(pos, i + 1 - pos); // take numeric value of part int cur = Int32.Parse(part); // if pos is 0 then just send numeric value // for next recursion if (pos == 0) getExprUtil(res, curExp + part, input, target, i + 1, cur, cur); // try all given binary operator for evaluation else { getExprUtil(res, curExp + "+" + part, input, target, i + 1, curVal + cur, cur); getExprUtil(res, curExp + "-" + part, input, target, i + 1, curVal - cur, -cur); getExprUtil(res, curExp + "*" + part, input, target, i + 1, curVal - last + last * cur, last * cur); } } } // Below method returns all possible expression // evaluating to target static List< string > getExprs( string input, int target) { List< string > res = new List< string >(); getExprUtil(res, "" , input, target, 0, 0, 0); return res; } // method to print result static void printResult(List< string > res) { for ( int i = 0; i < res.Count; i++) Console.Write(res[i] + " " ); Console.WriteLine(); } static void Main() { string input = "123" ; int target = 6; List< string > res = getExprs(input, target); printResult(res); input = "125" ; target = 7; res = getExprs(input, target); printResult(res); } } // This code is contributed by divyesh072019. |
Javascript
<script> // Javascript program to find all possible expression which // evaluate to target // Utility recursive method to generate all possible // expressions function getExprUtil(res, curExp, input, target, pos, curVal, last) { // true if whole input is processed with some // operators if (pos == input.length) { // if current value is equal to target //then only add to final solution // if question is : all possible o/p then just //push_back without condition if (curVal == target) res.push(curExp); return ; } // loop to put operator at all positions for (let i = pos; i < input.length; i++) { // ignoring case which start with 0 as they // are useless for evaluation if (i != pos && input[pos] == '0' ) break ; // take part of input from pos to i let part = input.substr(pos, i + 1 - pos); // take numeric value of part let cur = parseInt(part, 10); // if pos is 0 then just send numeric value // for next recursion if (pos == 0) getExprUtil(res, curExp + part, input, target, i + 1, cur, cur); // try all given binary operator for evaluation else { getExprUtil(res, curExp + "+" + part, input, target, i + 1, curVal + cur, cur); getExprUtil(res, curExp + "-" + part, input, target, i + 1, curVal - cur, -cur); getExprUtil(res, curExp + "*" + part, input, target, i + 1, curVal - last + last * cur, last * cur); } } } // Below method returns all possible expression // evaluating to target function getExprs(input, target) { let res = []; getExprUtil(res, "" , input, target, 0, 0, 0); return res; } // method to print result function printResult(res) { for (let i = 0; i < res.length; i++) document.write(res[i] + " " ); document.write( "</br>" ); } let input = "123" ; let target = 6; let res = getExprs(input, target); printResult(res); input = "125" ; target = 7; res = getExprs(input, target); printResult(res); // This code is contributed by decode2207. </script> |
输出:
1+2+3 1*2*3 1*2+5 12-5
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