[LeetCode] 241. Different Ways to Add Parentheses 添加括号的不同方式 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...
Leetcode: Different Ways to Add Parentheses Given a string of numbers and operators,returnall possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, - and *. Example1Input:"2-1-1". ((2-1)-1) = 0(2-(1-1)) = 2Out...
思路:类似unique binary tree, generate parentheses. 卡特兰数。 Runtime: 2 ms, faster than 84.52% of Java online submissions for Different Ways to Add Parentheses. Memory Usage: 38.6 MB, less than 18.35% of Java online submissions for Different Ways to Add Parentheses. class Solution { public L...
} else { left = diffWaysToCompute(input.substring(0, i));// 左右边所有可能的计算结果 } if (maps.containsKey(sright)) { right = maps.get(sright); } else { right = diffWaysToCompute(input.substring(i + 1)); } for (int l : left) { for (int r : right) { int res = 0; ...
Leetcode题目Different Ways to Add Parentheses的解题思路是什么? 如何使用递归解决Leetcode的Different Ways to Add Parentheses问题? Different Ways to Add Parentheses题目中,如何处理运算符的优先级? Given a string of numbers and operators, return all possible results from computing all the different possible...
leetcode [Divide and Conquer] No.241 Different Ways to Add Parentheses,程序员大本营,技术文章内容聚合第一站。
点击打开链接https://leetcode.com/problems/different-ways-to-add-parentheses/?tab=Description 给出一个表达式,求出所有可能的表达式的结果。 (1)对于加括号这个问题,是按照操作符分割来加的。所以每一次都是若干个分治过程,个数等于操作符数目。return case是没有操作符。这个有重叠子问题的特征,因...
原文地址:https://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...
Loading...leetcode.com/problems/different-ways-to-add-parentheses/ 中缀表达式,考虑一棵二叉树,每个内部结点是运算符,叶子结点是操作数。分治法,每一个运算符都有可能根结点。 classSolution{public:vector<int>diffWaysToCompute(stringinput){returncompute(input,0,input.length()-1);}private:vector<int...
publicList<Integer>diffWaysToCompute(Stringinput){if(input.length()==0){returnnewArrayList<>();}List<Integer>result=newArrayList<>();intnum=0;//考虑是全数字的情况intindex=0;while(index<input.length()&&!isOperation(input.charAt(index))){num=num*10+input.charAt(index)-'0';index++;}//将...