LeetCode -- Expression Add Operators

题目描述：


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 -> []


给定一串数，在数字中间插入操作符构成表达式，计算表达式，使得结果等于target。


思路：
由于要考虑多位数的情况，对于数组nums,需要对每一位nums[i]（其中i∈[0,n)）进行字符串截取: nums.Substr[i,i+1],nums.Substr[i,i+2]...nums.Substr[i,i+n-1]，然后对每个字串DFS。


一开始想DFS出所有表达式然后逐个计算的，可是表达式计算本身需要字符串遍历，使用栈和队列求解，因此这个方案不可取；
而是需要在遍历过程中立即计算，对于这个方案，要考虑到不同操作符+,-和*计算时的优先级问题。


1.对于+和-可以直接计算，然后从下一位作为起始进入DFS；
2.而对于*并且之前为+或-，需要恢复上一次+或-的计算值，先计算当前的*，然后再执行之前运算的逆运算。


其次，需要考虑当字符串长度大于1并且首字符为0的情况，直接返回。




参考连接：
http://www.sptzxb.com/oj/2015/09/21/LeetCode-Expression%20Add%20Operators/
http://bookshadow.com/weblog/2015/09/16/leetcode-expression-add-operators/




实现代码：

public class Solution {    public IList<string> AddOperators(string num, int target)     {        var result = new List<string>();    Dfs(target, num, 0, 0, '+', "", ref result);        return result;    }    private void Dfs(int target, string num, long current , long prevNum, char prevOp, string s, ref List<string> result){   if (num == "")   {   if(current == target){result.Add(s);}return;   }   for(var i = 1 ;i <= num.Length; i++){var str = num.Substring(0, i);if(str.Length > 1 && str[0] == '0'){return;}long n = long.Parse(str);var left = num.Substring(i, num.Length - i);if(s == ""){Dfs(target, left, n, n,'+', str, ref result);continue;}//Console.WriteLine(str + ","+index);Dfs(target, left, current + n, n,'+', s +"+"+str, ref result);Dfs(target, left, current - n, n,'-', s +"-"+str, ref result);// for '*' operator , execute reverse operation for previous operationif(prevOp == '+'){Dfs(target, left, current - prevNum + prevNum * n,  prevNum * n, prevOp, s +"*"+str, ref result);}else if(prevOp == '-'){Dfs(target, left, current + prevNum - prevNum * n,  prevNum * n, prevOp, s +"*"+str, ref result);}else{    Dfs(target, left, current * n, n, prevOp, s +"*"+str, ref result);}}}}