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Given a string that contains only digits 0-9 and a target value, return all possibilities to add binary operators (not unary) +, -, or * between the digits so they evaluate to the target value.
Examples:
"123", 6 -> ["1+2+3", "1*2*3"] "232", 8 -> ["2*3+2", "2+3*2"] "105", 5 -> ["1*0+5","10-5"] "00", 0 -> ["0+0", "0-0", "0*0"] "3456237490", 9191 -> []
class Solution { public: void dfs(string nums,int nums_size,int start_pos,vector<string>& res, int target,string &one_res,long long int total_res,long long int exp_res,char pre_op) { for(int i=start_pos;i<nums_size;i++){ if(i>start_pos && nums[start_pos]==‘0‘){ return; } long long int num = 0; for(int j=start_pos;j<=i;j++){ num = num*10+(nums[j]-‘0‘); } int one_res_size = one_res.size(); if(start_pos!=0){ one_res.push_back(pre_op); } one_res += nums.substr(start_pos,i-start_pos+1); long long int tmp_exp_res = exp_res; switch(pre_op){ case ‘*‘ : exp_res *= num; break; case ‘-‘ : exp_res = -num; break; case ‘+‘ : exp_res = num; break; } if(i+1==nums_size){ if(total_res+exp_res == target){ res.push_back(one_res); } }else{ dfs(nums,nums_size,i+1,res,target,one_res, total_res, exp_res, ‘*‘); dfs(nums,nums_size,i+1,res,target,one_res, total_res+exp_res, 0, ‘+‘); dfs(nums,nums_size,i+1,res,target,one_res, total_res+exp_res, 0, ‘-‘); } one_res.erase(one_res.begin()+one_res_size,one_res.end()); exp_res = tmp_exp_res; } } vector<string> addOperators(string nums, int target) { int nums_size = nums.size(); vector<string> res; string one_res; dfs(nums,nums_size,0,res,target,one_res,0,0,‘+‘); return res; } };
[string]Expression Add Operators
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原文地址：http://www.cnblogs.com/zengzy/p/5040575.html
