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.
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: []
Approach #1: DFS. [C++]
class Solution {
public:
vector<string> addOperators(string num, int target) {
vector<string> ans;
dfs(num, target, 0, "", 0, 0, &ans);
return ans;
}
private:
void dfs(const string& num, const int target,
int pos, const string& exp, long prev, long curr,
vector<string>* ans) {
if (pos == num.length()) {
if (curr == target) ans->push_back(exp);
return;
}
for (int l = 1; l <= num.length()-pos; ++l) {
string t = num.substr(pos, l);
if (t[0] == '0' && t.length() > 1) break;
long n = std::stol(t);
if (n > INT_MAX) break;
if (pos == 0) {
dfs(num, target, l, t, n, n, ans);
continue;
}
dfs(num, target, pos + l, exp + '+' + t, n, curr + n, ans);
dfs(num, target, pos + l, exp + '-' + t, -n, curr - n, ans);
dfs(num, target, pos + l, exp + '*' + t, prev * n, curr - prev + prev * n, ans);
}
}
};
Analysis:
http://zxi.mytechroad.com/blog/searching/leetcode-282-expression-add-operators/