<p>Given a string <code>expression</code> of numbers and operators, return <em>all possible results from computing all the different possible ways to group numbers and operators</em>. You may return the answer in <strong>any order</strong>.</p>

<p>The test cases are generated such that the output values fit in a 32-bit integer and the number of different results does not exceed <code>10<sup>4</sup></code>.</p>

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

<pre>
<strong>Input:</strong> expression = &quot;2-1-1&quot;
<strong>Output:</strong> [0,2]
<strong>Explanation:</strong>
((2-1)-1) = 0 
(2-(1-1)) = 2
</pre>

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

<pre>
<strong>Input:</strong> expression = &quot;2*3-4*5&quot;
<strong>Output:</strong> [-34,-14,-10,-10,10]
<strong>Explanation:</strong>
(2*(3-(4*5))) = -34 
((2*3)-(4*5)) = -14 
((2*(3-4))*5) = -10 
(2*((3-4)*5)) = -10 
(((2*3)-4)*5) = 10
</pre>

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

<ul>
	<li><code>1 &lt;= expression.length &lt;= 20</code></li>
	<li><code>expression</code> consists of digits and the operator <code>&#39;+&#39;</code>, <code>&#39;-&#39;</code>, and <code>&#39;*&#39;</code>.</li>
	<li>All the integer values in the input expression are in the range <code>[0, 99]</code>.</li>
</ul>

class Solution {
public:
    vector<int> diffWaysToCompute(string expression) {
        vector<int> data;
        vector<char> ops;
        istringstream ss(expression+"+");// "+"便于while()转换，不会用到
        int num = 0;
        char op = ' ';
        while(ss >> num && ss >> op) {
            data.push_back(num);
            ops.push_back(op);
        }
        int n = data.size();
        vector<vector<vector<int>>> dp(n, vector<vector<int>>(n, vector<int>()));
        for(int i = 0; i < n; ++i) { 
            for(int j = i; j >=0; --j) {//以i为结尾,j为开头的表达式
                if(i == j) dp[i][j].push_back(data[i]);
                else {
                    for(int k = j; k < i; ++k) {// 操作符位置。
                        for(auto left : dp[j][k]) {
                            for(auto right : dp[k+1][i]) {
                                int val = 0;
                                switch(ops[k]) {
                                    case '+': val = left + right;break;
                                    case '-': val = left - right;break;
                                    case '*': val = left * right;break;
                                }
                                dp[j][i].push_back(val);
                            }
                        }
                    }
                        
                }
            }
        }
        return dp[0][n-1];
    }
};
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