# [LeetCode] Expression Add Operators 表达式增加操作符

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.

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"]

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);
}
}
}
};

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