282. Expression Add Operators
问题描述:
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.
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: []
解题思路:
这道题可以想到使用dfs来做,但是在dfs的过程中遇到了麻烦:因为乘法的存在使得出现了优先级:乘法优先于加法和减法。
所以我们需要记录上一个操作数:
e.g.1 a+b*c 上一个操作数就是 b*c
e.g.2 b*c + a上一个操作数就是a
e.g. 3 b*c -a 上一个操作数就是-a
这里参考了Hcisly的解法(在下面回复中)
void dfs(string& s, int target, int pos, long cv, long pv, string r, vector<string>& res)
s : 给出的数字
target: 目标值
pos: 起始位置
cv: 当前的累计值
pv: 上一个操作数的值
注意pv的符号!
代码:
class Solution { public: vector<string> addOperators(string num, int target) { vector<string> ret; dfs(num, target, 0, 0, 0, "", ret); return ret; } void dfs(string& s, int target, int pos, long cv, long pv, string r, vector<string>& res) { if (pos == s.size() && cv == target) { res.push_back(r); return; } for (int i = 1; i <= s.size() - pos; i++) { string t = s.substr(pos, i); if (i > 1 && t[0] == '0') break; // preceding long n = stol(t); if (pos == 0) { dfs(s, target, i, n, n, t, res); continue; } dfs(s, target, pos+i, cv+n, n, r+"+"+t, res); dfs(s, target, pos+i, cv-n, -n, r+"-"+t, res); dfs(s, target, pos+i, cv-pv+pv*n, pv*n, r+"*"+t, res); } } };