LeetCode282. Expression Add Operators

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: []

思路：

我们这里参看了LeetCode提供的解答：here。当然，仅仅靠上面的解答完成的CPP程序最后一个实例会超时，所以我们做了一些改进，主要是在traceback的时候，我们直接用赋值的方式替换了“还原的方式”。

class Solution {
public:
	vector addOperators(string num, int target) {
		this->num = num;
		this->target = target;
		recurse(0, 0, 0, 0, string());
		return res;
	}
private:
	vector res;
	string num;
	int target;
	void recurse(int index, long pre_num, long current_num, long val, string candidate) {
		if (index == num.size()) {
			if (val == target && current_num == 0) res.push_back(candidate.substr(1));
			return;
		}
		current_num = current_num * 10 + num[index] - '0';
		string curr_str = to_string(current_num);
		if (current_num > 0) recurse(index + 1, pre_num, current_num, val, candidate);

		int n = candidate.size();
		candidate.push_back('+'); candidate += curr_str;
		recurse(index + 1, current_num, 0, val + current_num, candidate);

		if (n > 0) {
			candidate[n] = '-';
			recurse(index + 1, -current_num, 0, val - current_num, candidate);

			candidate[n] = '*';
			recurse(index + 1, current_num * pre_num, 0, val - pre_num + pre_num * current_num, candidate);
		}
	}
};