282. Expression Add Operators_给定一个包含0-9的字符串和一个目标值-优快云博客

本文介绍了一种通过递归回溯算法解决给定数字串中添加运算符使其等于目标值的问题的方法。通过C++实现，详细展示了如何遍历所有可能的运算组合。

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题目：

Given a string that contains only digits 0-9 and a target value, return all possibilities to add binary operators (not unary) +, -, or *between the digits so they evaluate to the target value.

Examples:

"123", 6 -> ["1+2+3", "1*2*3"] 
"232", 8 -> ["2*3+2", "2+3*2"]
"105", 5 -> ["1*0+5","10-5"]
"00", 0 -> ["0+0", "0-0", "0*0"]
"3456237490", 9191 -> []

Credits:
Special thanks to @davidtan1890 for adding this problem and creating all test cases.

翻译：、

给定一个仅包含数字0-9和目标值的字符串，返回在数字之间添加二进制运算符（不是一元）+， - 或*的所有可能性，以便它们计算目标值

例子：
“123”，6 - > [“1 + 2 + 3”，“1 * 2 * 3”]
“232”，8 - > [“2 * 3 + 2”，“2 + 3 * 2”]
“105”，5 - > [“1 * 0 + 5”，“10-5”]
“00”，0 - > [“0 + 0”，“0-0”，“0 * 0”]
“3456237490”，9191 - > []
学分：
特别感谢@ davidtan1890添加此问题并创建所有测试用例。

解题代码：

class Solution {
private:
		void f(vector<string> &strs, int target, string cur,string nums, long cv, long pv) {
			if(nums.length() == 0) {
				if(target == cv) strs.push_back(cur);
				return ;
			}
			else {
				for(int i = 1; i<= nums.length();i++) {
					string num = nums.substr(0,i);
					if(num.length() > 1 && num[0] == '0') return ;
				//	char* flag = NULL;
				//	char tem[i+1];
				//	strcpy(tem,num.c_str());
				//	cout << tem <<endl;
				//	long l = strtol(tem,&flag,10);
				long l  = stol(num);
				//	if(*flag != '\0') continue;
				//	cout << l<<endl; 
					string nextNums = nums.substr(i);
					if(cur.length()!=0) {
						f(strs,target,cur+'+'+num,nextNums,cv+l,l);
						f(strs,target,cur+'-'+num,nextNums,cv-l,-l);
						f(strs,target,cur+'*'+num,nextNums,cv-pv+pv*l,pv*l);
					}	
					else {
						f(strs,target,num,nextNums,l,l);
					}
				}
			}
		}
public:
    vector<string> addOperators(string num, int target) {
		vector<string> strs;
		strs.clear();
		if(num.empty()) return strs;
		 f(strs,target,"",num,0,0);
		 return strs;
    }
};

解题状态：

class Solution {
private:
		void f(vector<string> &strs, int target, string cur,string nums, long cv, long pv) {
			if(nums.length() == 0) {
				if(target == cv) strs.push_back(cur);
				return ;
			}
			else {
				for(int i = 1; i<= nums.length();i++) {
					string num = nums.substr(0,i);
					if(num.length() > 1 && num[0] == '0') return ;
				//	char* flag = NULL;
				//	char tem[i+1];
				//	strcpy(tem,num.c_str());
				//	cout << tem <<endl;
				//	long l = strtol(tem,&flag,10);
				long l  = stol(num);
				//	if(*flag != '\0') continue;
				//	cout << l<<endl; 
					string nextNums = nums.substr(i);
					if(cur.length()!=0) {
						f(strs,target,cur+'+'+num,nextNums,cv+l,l);
						f(strs,target,cur+'-'+num,nextNums,cv-l,-l);
						f(strs,target,cur+'*'+num,nextNums,cv-pv+pv*l,pv*l);
					}	
					else {
						f(strs,target,num,nextNums,l,l);
					}
				}
			}
		}
public:
    vector<string> addOperators(string num, int target) {
		vector<string> strs;
		strs.clear();
		if(num.empty()) return strs;
		 f(strs,target,"",num,0,0);
		 return strs;
    }
};