Leetcode 282 给表达式添加运算符 +-*运算暴搜处理

+，-，*有一种特别直接的处理方法是直接算，保存一个答案ans，pre，遇到运算符就算，遇到乘法的时候就先把之前算的减掉ans-pre+pre*cur，就这样处理即可。

typedef long long LL;
class Solution {
public:
    vector<string> addOperators(string num, int target) {
        string expression;
        helper(num,target,0,0,0,expression);
        return res;
    }
    vector<string> res;
    void helper(string num, int target, int index, LL ans, LL pre, string expression){
        if(index==num.size()){
            if(target==(int)ans) res.push_back(expression);
            return;
        }
        for(int i=index;i<num.size();i++){
            string s = num.substr(index, i - index + 1);       // 从index向后取出对应长度的新字符串
            if(s.size() > 1 && s[0] == '0') return;            // 如果第一位是0的两位数，那么跳过
            long cur = stol(s);
            if(index==0) helper(num,target,i+1,ans+cur,cur,expression+s);
            else{
                helper(num,target,i+1,ans+cur,cur,expression+"+"+s);
                helper(num,target,i+1,ans-cur,-cur,expression+"-"+s);
                helper(num,target,i+1,ans- pre + pre * cur, pre*cur,expression+"*"+s);
            }
        }
    }
};

将传值改成传引用，并加入回溯处理

typedef long long LL;
class Solution {
public:
    vector<string> addOperators(string num, int target) {
        string expression;
        helper(num,target,0,0,0,expression);
        return res;
    }
    vector<string> res;
    void helper(string num, int target, int index, LL ans, LL pre, string expression){
        if(index==num.size()){
            if(target==(int)ans) res.push_back(expression);
            return;
        }
        for(int i=index;i<num.size();i++){
            string s = num.substr(index, i - index + 1);       // 从index向后取出对应长度的新字符串
            if(s.size() > 1 && s[0] == '0') return;            // 如果第一位是0的两位数，那么跳过
            long cur = stol(s);
            if(index==0) helper(num,target,i+1,ans+cur,cur,expression+s);
            else{
                helper(num,target,i+1,ans+cur,cur,expression+"+"+s);
                helper(num,target,i+1,ans-cur,-cur,expression+"-"+s);
                helper(num,target,i+1,ans- pre + pre * cur, pre*cur,expression+"*"+s);
            }
        }
    }
};