241. 为运算表达式设计优先级

English Version

题目描述

给你一个由数字和运算符组成的字符串 expression ，按不同优先级组合数字和运算符，计算并返回所有可能组合的结果。你可以 按任意顺序 返回答案。

生成的测试用例满足其对应输出值符合 32 位整数范围，不同结果的数量不超过 104 。

 

示例 1：

输入：expression = "2-1-1"
输出：[0,2]
解释：
((2-1)-1) = 0 
(2-(1-1)) = 2

示例 2：

输入：expression = "2*3-4*5"
输出：[-34,-14,-10,-10,10]
解释：
(2*(3-(4*5))) = -34 
((2*3)-(4*5)) = -14 
((2*(3-4))*5) = -10 
(2*((3-4)*5)) = -10 
(((2*3)-4)*5) = 10

 

提示：

• 1 <= expression.length <= 20
• expression 由数字和算符 '+'、'-' 和 '*' 组成。
• 输入表达式中的所有整数值在范围 [0, 99] 

解法

方法一：记忆化搜索

Python3

class Solution:
    def diffWaysToCompute(self, expression: str) -> List[int]:
        @cache
        def dfs(exp):
            if exp.isdigit():
                return [int(exp)]
            ans = []
            for i, c in enumerate(exp):
                if c in '-+*':
                    left, right = dfs(exp[:i]), dfs(exp[i + 1 :])
                    for a in left:
                        for b in right:
                            if c == '-':
                                ans.append(a - b)
                            elif c == '+':
                                ans.append(a + b)
                            else:
                                ans.append(a * b)
            return ans

        return dfs(expression)

Java

class Solution {
    private static Map<String, List<Integer>> memo = new HashMap<>();

    public List<Integer> diffWaysToCompute(String expression) {
        return dfs(expression);
    }

    private List<Integer> dfs(String exp) {
        if (memo.containsKey(exp)) {
            return memo.get(exp);
        }
        List<Integer> ans = new ArrayList<>();
        if (exp.length() < 3) {
            ans.add(Integer.parseInt(exp));
            return ans;
        }
        for (int i = 0; i < exp.length(); ++i) {
            char c = exp.charAt(i);
            if (c == '-' || c == '+' || c == '*') {
                List<Integer> left = dfs(exp.substring(0, i));
                List<Integer> right = dfs(exp.substring(i + 1));
                for (int a : left) {
                    for (int b : right) {
                        if (c == '-') {
                            ans.add(a - b);
                        } else if (c == '+') {
                            ans.add(a + b);
                        } else {
                            ans.add(a * b);
                        }
                    }
                }
            }
        }
        memo.put(exp, ans);
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> diffWaysToCompute(string expression) {
        return dfs(expression);
    }

    vector<int> dfs(string exp) {
        if (memo.count(exp)) return memo[exp];
        if (exp.size() < 3) return {stoi(exp)};
        vector<int> ans;
        int n = exp.size();
        for (int i = 0; i < n; ++i) {
            char c = exp[i];
            if (c == '-' || c == '+' || c == '*') {
                vector<int> left = dfs(exp.substr(0, i));
                vector<int> right = dfs(exp.substr(i + 1, n - i - 1));
                for (int& a : left) {
                    for (int& b : right) {
                        if (c == '-')
                            ans.push_back(a - b);
                        else if (c == '+')
                            ans.push_back(a + b);
                        else
                            ans.push_back(a * b);
                    }
                }
            }
        }
        memo[exp] = ans;
        return ans;
    }

private:
    unordered_map<string, vector<int>> memo;
};

Go

var memo = map[string][]int{}

func diffWaysToCompute(expression string) []int {
	return dfs(expression)
}

func dfs(exp string) []int {
	if v, ok := memo[exp]; ok {
		return v
	}
	if len(exp) < 3 {
		v, _ := strconv.Atoi(exp)
		return []int{v}
	}
	ans := []int{}
	for i, c := range exp {
		if c == '-' || c == '+' || c == '*' {
			left, right := dfs(exp[:i]), dfs(exp[i+1:])
			for _, a := range left {
				for _, b := range right {
					if c == '-' {
						ans = append(ans, a-b)
					} else if c == '+' {
						ans = append(ans, a+b)
					} else {
						ans = append(ans, a*b)
					}
				}
			}
		}
	}
	memo[exp] = ans
	return ans
}

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