如何用Python编写LeetCode241.不同加括号方式求解算法的笔记?

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2026-09-24 02:40:06
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本文共计262个文字,预计阅读时间需要2分钟。

如何用Python编写LeetCode241.不同加括号方式求解算法的笔记?

题目:https://leetcode.com/problems/different-ways-to-add-parentheses/description/+给定一个数字和运算符的字符串,返回计算所有可能的分组方式所得的所有结果。+

题目:leetcode.com/problems/different-ways-to-add-parentheses/description/

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, - and *.

Example 1:

Input: “2-1-1”.

((2-1)-1) = 0(2-(1-1)) = 2

Output: [0, 2]

Example 2:

如何用Python编写LeetCode241.不同加括号方式求解算法的笔记?

Input: “23-45”

(2*(3-(45))) = -34((23)-(45)) = -14((2(3-4))5) = -10(2((3-4)5)) = -10(((23)-4)*5) = 10

Output: [-34, -14, -10, -10, 10]

分析:递归,笛卡尔积代码:

class Solution: def diffWaysToCompute(self, input): ops = {'+': lambda x, y: x + y, '-': lambda x, y: x - y, '*': lambda x, y: x * y} def ways(s): ans = [] for i in range(len(s)): if s[i] in "+-*": ans += [ops[s[i]](l, r) for l, r in itertools.product(ways(s[0:i]), ways(s[i+1:]))] if not ans: ans.append(int(s)) return ans return ways(input)

代码转自:zxi.mytechroad.com/blog/leetcode/leetcode-241-different-ways-to-add-parentheses/

本文共计262个文字,预计阅读时间需要2分钟。

如何用Python编写LeetCode241.不同加括号方式求解算法的笔记?

题目:https://leetcode.com/problems/different-ways-to-add-parentheses/description/+给定一个数字和运算符的字符串,返回计算所有可能的分组方式所得的所有结果。+

题目:leetcode.com/problems/different-ways-to-add-parentheses/description/

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, - and *.

Example 1:

Input: “2-1-1”.

((2-1)-1) = 0(2-(1-1)) = 2

Output: [0, 2]

Example 2:

如何用Python编写LeetCode241.不同加括号方式求解算法的笔记?

Input: “23-45”

(2*(3-(45))) = -34((23)-(45)) = -14((2(3-4))5) = -10(2((3-4)5)) = -10(((23)-4)*5) = 10

Output: [-34, -14, -10, -10, 10]

分析:递归,笛卡尔积代码:

class Solution: def diffWaysToCompute(self, input): ops = {'+': lambda x, y: x + y, '-': lambda x, y: x - y, '*': lambda x, y: x * y} def ways(s): ans = [] for i in range(len(s)): if s[i] in "+-*": ans += [ops[s[i]](l, r) for l, r in itertools.product(ways(s[0:i]), ways(s[i+1:]))] if not ans: ans.append(int(s)) return ans return ways(input)

代码转自:zxi.mytechroad.com/blog/leetcode/leetcode-241-different-ways-to-add-parentheses/