Leetcode | Expression Add Operators
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原题链接:https://leetcode.com/problems/expression-add-operators
题目内容如下所示:
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 -> []
题目的大概意思就是给出一个包含0-9的字符串,然后向其中任意添加一些二元操作符,包括+、-和*,添加后会得到一个计算表达式,算出来的结果要是后面给出的数字target,所以是要我们输出所有可能的表达式的集合。
这道题刚着手时并不简单,虽然题意看起来很容易理解,但仔细想想其实还挺复杂的,因为它可以是相邻数字组在一起当成一个更大的数字,而不是固定地在字符串中每个数字中间插一个操作符,所以这其实可以看成一个动态规划的问题,我们可以先从它的子问题入手,在这道题就相当于是子字符串,然后找出子串和原串之间的联系,这里其实可以用dfs的思想,通过递归遍历的方式逐渐找出问题的所有解。
最后给出代码如下所示:
class Solution {private: void dfs(std::vector<string>& res, const string& num, const int target, string cur, int pos, const long cv, const long pv, const char op) { if (pos == num.size() && cv == target) { res.push_back(cur); } else { for (int i=pos+1; i<=num.size(); i++) { string t = num.substr(pos, i-pos); long now = stol(t); if (to_string(now).size() != t.size()) continue; dfs(res, num, target, cur+'+'+t, i, cv+now, now, '+'); dfs(res, num, target, cur+'-'+t, i, cv-now, now, '-'); dfs(res, num, target, cur+'*'+t, i, (op == '-') ? cv+pv - pv*now : ((op == '+') ? cv-pv + pv*now : pv*now), pv*now, op); } } }public: vector<string> addOperators(string num, int target) { vector<string> res; if (num.empty()) return res; for (int i=1; i<=num.size(); i++) { string s = num.substr(0, i); long cur = stol(s); if (to_string(cur).size() != s.size()) continue; dfs(res, num, target, s, i, cur, cur, '#'); } return res; }};
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