<p>Given a string <code>num</code> that contains only digits and an integer <code>target</code>, return <em><strong>all possibilities</strong> to insert the binary operators </em><code>&#39;+&#39;</code><em>, </em><code>&#39;-&#39;</code><em>, and/or </em><code>&#39;*&#39;</code><em> between the digits of </em><code>num</code><em> so that the resultant expression evaluates to the </em><code>target</code><em> value</em>.</p>

<p>Note that operands in the returned expressions <strong>should not</strong> contain leading zeros.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> num = &quot;123&quot;, target = 6
<strong>Output:</strong> [&quot;1*2*3&quot;,&quot;1+2+3&quot;]
<strong>Explanation:</strong> Both &quot;1*2*3&quot; and &quot;1+2+3&quot; evaluate to 6.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> num = &quot;232&quot;, target = 8
<strong>Output:</strong> [&quot;2*3+2&quot;,&quot;2+3*2&quot;]
<strong>Explanation:</strong> Both &quot;2*3+2&quot; and &quot;2+3*2&quot; evaluate to 8.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> num = &quot;3456237490&quot;, target = 9191
<strong>Output:</strong> []
<strong>Explanation:</strong> There are no expressions that can be created from &quot;3456237490&quot; to evaluate to 9191.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= num.length &lt;= 10</code></li>
	<li><code>num</code> consists of only digits.</li>
	<li><code>-2<sup>31</sup> &lt;= target &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution {
public:
    vector<string> result;
    int target;

    void addResult(vector<long long> &digits, vector<char> &ops){
        string s = "";
        for(int i=0; i < digits.size()-1; i++){
            s += to_string(digits[i]) + ops[i];
        }
        s += to_string(digits.back());
        result.push_back(s);
    }

    void addOps(vector<long long> &digits, vector<char> &ops, vector<long long> &st, int start){
        if (start == digits.size()){
            
            if (accumulate(st.begin(), st.end(), 0LL) == target){
                addResult(digits, ops);
            }
        } else if (st.empty()) {
            st.push_back(digits[start]);
            addOps(digits, ops, st, start+1);
        } else {
            ops.push_back('+');
            st.push_back(digits[start]);
            addOps(digits, ops, st, start+1);
            ops.pop_back();
            st.pop_back();

            ops.push_back('-');
            st.push_back(-digits[start]);
            addOps(digits, ops, st, start+1);
            ops.pop_back();
            st.pop_back();

            ops.push_back('*');
            auto tmp = st.back();
            st.back() *= digits[start];
            addOps(digits, ops, st, start+1);
            st.back() = tmp; // 必須撤銷之前的乘法操作
            ops.pop_back();
        }
    }

    void parseNum(string &num, vector<long long> &digits, int cr){
        if (cr == num.size()){
            vector<char> ops;
            vector<long long> st;
            addOps(digits, ops, st, 0); 
            return;
        }
        
        digits.push_back(0); 
        for(int i=cr; i < num.size(); i++){
            digits.back() = digits.back()*10 + num[i] - '0';
            parseNum(num, digits, i+1);
            if (num[cr]=='0') {
                digits.pop_back(); 
                return ;
            }
        }
        digits.pop_back(); 
    }

    vector<string> addOperators(string num, int target) {
        this->target = target; 
        vector<long long> digits;
        parseNum(num, digits, 0);
        return result;
    }
};
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