Given a string that contains only digits 0-9 and a target value, return all possibilities to add binaryoperators (not unary) +, -, or * between the digits so they evaluate to the target value.
Example 1:
Input: num = "123", target = 6
Output: ["1+2+3", "1*2*3"]
Example 2:
Input: num = "232", target = 8
Output: ["2*3+2", "2+3*2"]
Example 3:
Input: num = "105", target = 5
Output: ["1*0+5","10-5"]
Example 4:
Input: num = "00", target = 0
Output: ["0+0", "0-0", "0*0"]
Example 5:
Input: num = "3456237490", target = 9191
Output: []
Credits:
Special thanks to @davidtan1890 for adding this problem and creating all test cases.
这道题给了我们一个只由数字组成的字符串，让我们再其中添加+,-或*号来形成一个表达式，该表达式的计算和为给定了target值，让我们找出所有符合要求的表达式来。看了题目中的例子1和2，很容易让人误以为是必须拆成个位数字，其实不是的，比如例子3中的 "105", 5能返回"10-5"，说明连着的数字也可以。如果非要在过往的题中找一道相似的题，我觉得跟 Combination Sum II 很类似。不过这道题要更复杂麻烦一些。还是用递归来解题，我们需要两个变量diff和curNum，一个用来记录将要变化的值，另一个是当前运算后的值，而且它们都需要用 long 型的，因为字符串转为int型很容易溢出，所以我们用长整型。对于加和减，diff就是即将要加上的数和即将要减去的数的负值，而对于乘来说稍有些复杂，此时的diff应该是上一次的变化的diff乘以即将要乘上的数，有点不好理解，那我们来举个例子，比如 2+3*2，即将要运算到乘以2的时候，上次循环的 curNum = 5, diff = 3, 而如果我们要算这个乘2的时候，新的变化值diff应为 3*2=6，而我们要把之前+3操作的结果去掉，再加上新的diff，即 (5-3)+6=8，即为新表达式 2+3*2 的值，有点难理解，大家自己一步一步推算吧。
还有一点需要注意的是，如果输入为"000",0的话，容易出现以下的错误：
Wrong：["0+0+0","0+0-0","0+0*0","0-0+0","0-0-0","0-0*0","0*0+0","0*0-0","0*0*0","0+00","0-00","0*00","00+0","00-0","00*0","000"]
Correct：["0*0*0","0*0+0","0*0-0","0+0*0","0+0+0","0+0-0","0-0*0","0-0+0","0-0-0"]
我们可以看到错误的结果中有0开头的字符串出现，明显这不是数字，所以我们要去掉这些情况，过滤方法也很简单，我们只要判断长度大于1且首字符是‘0’的字符串，将其滤去即可，参见代码如下：
class Solution { public: vector<string> addOperators(string num, int target) { vector<string> res; helper(num, target, 0, 0, "", res); return res; } void helper(string num, int target, long diff, long curNum, string out, vector<string>& res) { if (num.size() == 0 && curNum == target) { res.push_back(out); return; } for (int i = 1; i <= num.size(); ++i) { string cur = num.substr(0, i); if (cur.size() > 1 && cur[0] == '0') return; string next = num.substr(i); if (out.size() > 0) { helper(next, target, stoll(cur), curNum + stoll(cur), out + "+" + cur, res); helper(next, target, -stoll(cur), curNum - stoll(cur), out + "-" + cur, res); helper(next, target, diff * stoll(cur), (curNum - diff) + diff * stoll(cur), out + "*" + cur, res); } else { helper(next, target, stoll(cur), stoll(cur), cur, res); } } } };
类似题目：
Evaluate Reverse Polish Notation
Different Ways to Add Parentheses
参考资料：
https://leetcode.com/problems/expression-add-operators/
https://leetcode.com/problems/expression-add-operators/discuss/71971/Accepted-C%2B%2B-Solution