Leetcode 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: []

Accepted

89,424

Submissions

257,426

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这题目并不像24点一样可以加括号，所以递推时加减过的表达式不能再去乘，这点一定要明确，写得过程中思路很清晰，但是坑不少，自己的解法用了大量的set，所以会偏慢，以下是自己的解法：

class Solution:
    def addOperators(self, num, target):
        l,res = len(num),[]
        if (l == 0):
            return res
        exps = [[set() for j in range(l)] for i in range(l)]
        for i in range(l):
            exps[i][i].add((num[i],int(num[i]),0))

        for w in range(2,l+1):
            for i in range(0,l+1-w):
                j = i+w-1
                for mid in range(i,j):
                    lset,rset = exps[i][mid],exps[mid+1][j]
                    if (num[i] != '0'): #bug1: forget
                        exps[i][j].add((num[i:j+1],int(num[i:j+1]),0))
                    ops = ['+','-','*'] if i == 0 else ['*']

                    for ll in lset:
                        for rr in rset:
                            for op in ops:
                                if (op == '+'):
                                    exps[i][j].add((ll[0]+'+'+rr[0], ll[1]+rr[1], 1))
                                elif (op == '-'):
                                    exps[i][j].add((ll[0]+'-'+rr[0], ll[1]-rr[1], 1))
                                elif (op == '*' and ll[2] == 0): #bug2: miss ll[2] == 0
                                    exps[i][j].add((ll[0]+'*'+rr[0], ll[1]*rr[1], 0))
        for exp in exps[0][l-1]:
            if (exp[1] == target):
                res.append(exp[0])
        return res
s = Solution()
print(s.addOperators(num = "00", target = 0))

从Discussion看到的别人的解法：

class Solution {
public:
    vector<string> addOperators(string num, int target) {
        vector<string> res;
        dfs(num, target, 0, 0, 0, "", res);
        return res;
    }
    
    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') continue; // 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);
        }
    }
};