标签：operator 回溯 bst 基本 dig 不能 str express span
题目：
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 -> []
题解：
想了半个小时还是想不出来，卡在乘上了，看了别人的代码，感觉这个处理很巧妙，基本思路和回溯递归差不多
Soution 1 ()
class Solution { public: void helper(string num, vector<string>& res, string s, int target, int pos, long cur, long pre) { if(pos == num.size()) { if(cur == target) res.push_back(s); return; } for(int i=pos; i<num.size(); i++) { //首字符为0且长度大于1，那么这个字符串不能代表数字 if(num[pos] == ‘0‘ && i>pos) break; string str = num.substr(pos, i-pos+1); long val = stol(str); if(pos == 0) helper(num, res, s+str, target, i+1, val, val); else { helper(num, res, s+‘+‘+str, target, i+1, cur+val, val); helper(num, res, s+‘-‘+str, target, i+1, cur-val, -val); helper(num, res, s+‘*‘+str, target, i+1, cur-pre+pre*val, pre*val); } } } vector<string> addOperators(string num, int target) { vector<string> res; if(num.size() == 0) return res; helper(num, res, "", target, 0, 0, 0); return res; } };
【LeetCode】282. Expression Add Operators
标签：operator 回溯 bst 基本 dig 不能 str express span
原文地址：http://www.cnblogs.com/Atanisi/p/6814841.html
