282. Expression Add Operators
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.
Example 1:
Input: num = "123", target = 6 Output: ["1+2+3", "1*2*3"]
Example 2:
Input: num = "232", target = 8 Output: ["2*3+2", "2+3*2"]
Example 3:
Input: num = "105", target = 5 Output: ["1*0+5","10-5"]
Example 4:
Input: num = "00", target = 0 Output: ["0+0", "0-0", "0*0"]
Example 5:
Input: num = "3456237490", target = 9191 Output: []
Approach #1: DFS. [C++]
class Solution { public: vector<string> addOperators(string num, int target) { vector<string> ans; dfs(num, target, 0, "", 0, 0, &ans); return ans; } private: void dfs(const string& num, const int target, int pos, const string& exp, long prev, long curr, vector<string>* ans) { if (pos == num.length()) { if (curr == target) ans->push_back(exp); return; } for (int l = 1; l <= num.length()-pos; ++l) { string t = num.substr(pos, l); if (t[0] == '0' && t.length() > 1) break; long n = std::stol(t); if (n > INT_MAX) break; if (pos == 0) { dfs(num, target, l, t, n, n, ans); continue; } dfs(num, target, pos + l, exp + '+' + t, n, curr + n, ans); dfs(num, target, pos + l, exp + '-' + t, -n, curr - n, ans); dfs(num, target, pos + l, exp + '*' + t, prev * n, curr - prev + prev * n, ans); } } };
Analysis:
http://zxi.mytechroad.com/blog/searching/leetcode-282-expression-add-operators/
永远渴望,大智若愚(stay hungry, stay foolish)