LINTCODE——Add Operators

LINTCODE——Add Operators

思路：用回溯法去试，注意两个特殊情况：
一、数字越界，如果大于INT_MAX则溢出，不过给的数据没有这个，我把越界的检查删除了也AC了；
二、数字的有效性，即0987不是有效数字，这种情况要排除，排除方法也很简单，例如已知字符串str =“0987”，把str转换成long long 或者long型得到curNum，再次把curNum，to_string一下，比较两者字符串的长度就OK了；

class Solution {private:    vector<string> ans;public:    /*     * @param num: a string contains only digits 0-9     * @param target: An integer     * @return: return all possibilities     */    vector<string> addOperators(string &num, int target) {        // write your code here        for(int i = 1; i <= num.size(); i++)        {            string curStr = num.substr(0,i);            long long curNum = stol(curStr);            if(to_string(curNum).size() != i)                continue;            dfs(num, target, curStr, i, curNum, curNum, '#');        }        return ans;    }    //curStr为当前字符串，即加了op运算符的字符串    //index为num的索引，判断是否结束递归    //cv为当前字符串的值；    //pv为前一块运算的值    //op为上一个运算符的符号，初试为‘#’    void dfs(const string &num, const int &target, string curStr, int index, long long cv, long long pv, char op)    {        if(index == num.size() && cv == target)        {            ans.push_back(curStr);            return;        }        for(int i = index+1; i <= num.size(); i++)        {            string s = num.substr(index,i-index);            long long curNum = stol(s);            if(to_string(curNum).size() != s.size() )                continue;            dfs(num, target, curStr+'+'+s, i, cv+curNum, curNum, '+');            dfs(num, target, curStr+'-'+s, i, cv-curNum, curNum, '-');            dfs(num, target, curStr+'*'+s, i,(op=='+')?(cv-pv+pv*curNum):((op=='-')?cv+pv-pv*curNum: cv*curNum), curNum*pv, op);        }        return;    }};