Expression Add Operators

此算法属于分治算法范畴，找到合理的分治策略即可：

vector<string> addOperators(string num, int target) {        vector<string> result;        operators_(num, 0, 0, "", target, result);        return result;    }    // 例如：数字串abc，想得到a + b * c的计算形式;第一次调用后out为a;第二次调用后out为a + b;第三次调用由于此时*的优先级更高,因此需要先计算b*c，然后再计算a+b*c;函数中diff为上次的变化量，针对加法，diff为正数；针对减法，diff为负数；    // 另外需要注意的细节是，中间分隔出来首位为0的数字，之多为1位；例如105，可以分割为1*0+5;1005不能分割为1*00+5;    void operators_(string& num, long long diff, long long curNum, string out, long long target, vector<string>& result){        if(num.empty() && curNum == target)            result.push_back(out);        for(int k = 1; k <= num.size(); ++k){            string cur = num.substr(0, k);            if(cur.size() > 1 && cur[0] == '0')                return ;            string second = num.substr(k);            if(!out.empty()){                operators_(second, stoll(cur), curNum + stoll(cur), out + "+" + cur, target, result);                operators_(second, -stoll(cur), curNum - stoll(cur), out + "-" + cur, target, result);                operators_(second, diff * stoll(cur), curNum - diff + diff * stoll(cur), out + "*" + cur, target, result);            }else{                operators_(second, stoll(cur), stoll(cur), cur, target, result);            }        }       }

文章思路参考

http://www.cnblogs.com/grandyang/p/4814506.html