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: …

Different Ways to Add Parentheses 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". ((…

lc 241 Different Ways to Add Parentheses 241 Different Ways to Add Parentheses 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…

96. Unique Binary Search Trees https://www.cnblogs.com/grandyang/p/4299608.html 3由dp[1]*dp[1].dp[0]*dp[2].dp[2]*dp[0]相加而成 从2开始 class Solution { public: int numTrees(int n) { vector<); dp[] = ; dp[] = ; ;i <= n;i++){ ;j < i;j++){ dp[i] += dp[j] *…

241. Different Ways to Add Parentheses https://leetcode.com/problems/different-ways-to-add-parentheses/ 思路就是:首先找到以运算符为根节点,分别计算左子串和右子串的所有结果的集合,然后依次进行组合计算.参考博客http://www.cnblogs.com/ganganloveu/p/4681439.html. 自己的思路错了,直接用两边只用了一个整数去接收左右子串的计算值!! #include…

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" Output: [0, 2] Explanation: ((2-1)-…

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: …

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" Output: [0, 2] Explanation: ((2-1)-…

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: …

https://leetcode.com/problems/different-ways-to-add-parentheses/ 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 *. E…

Problem: 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)) =…

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: …

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 方法一:递归构建所有表达式 方法二:分而治之 日期 题目地址:https://leetcode.com/problems/different-ways-to-add-parentheses/description/ 题目描述 Given a string of numbers and operators, return all possible re…

题目: 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" Output: [0, 2] Explanation: ((2…

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:…

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:…

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" Output: [0, 2] Explanation: ((2-1)-…

题目链接:https://leetcode.com/submissions/detail/86532557/ 算法类型:分治法 题目分析:计算表达式的所有结果可能性 代码实现: class Solution(object): def diffWaysToCompute(self, input): """ :type input: str :rtype: List[int] """ def dfs(s, cache) : ops = {'+':la…

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: …

题目: 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 Out…

给定一个含有数字和运算符的字符串,为表达式添加括号,改变其运算优先级以求出不同的结果.你需要给出所有可能的组合的结果.有效的运算符号包含 +, - 以及 * . 示例 1: 输入: "2-1-1" 输出: [0, 2] 解释: ((2-1)-1) = 0 (2-(1-1)) = 2 示例 2: 输入: "2*3-4*5" 输出: [-34, -14, -10, -10, 10] 解释: (2*(3-(4*5))) = -34 ((2*3)-(4*5)) = -14…

UNIX / Linux: 2 Ways to Add Swap Space Using dd, mkswap and swapon by RAMESH NATARAJAN on AUGUST 18, 2010   Question: I would like to add more swap space to my Linux system. Can you explain with clear examples on how to increase the swap space? Answe…

目录 题目链接 注意点 解法 小结 题目链接 Generate Parentheses - LeetCode 注意点 解法 解法一:递归.当left>right的时候返回(为了防止出现 )( ) class Solution { public: void recursion(int left,int right,string str,vector<string> &ret) { if(left > right) return; else if(left== 0&&a…

目录 题目链接 注意点 解法 小结 题目链接 Valid Parentheses - LeetCode 注意点 考虑输入为空的情况 解法 解法一:如果是'('.'{'.'['这三者就入栈,否则就判断栈是否为空和栈顶括号是否与之匹配.注意两个判断顺序不可以颠倒,不然会runtime error.时间复杂度为O(n) class Solution { public: bool isValid(string s) { stack<char> stk; for(auto &ch:s) { if…

1- 问题描述 Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "(…

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = 3, a solution set is: "((()))", "(()())", "(())()", "()(())", "()()()" 题目大意:给一…

回溯法 百度百科:回溯法(探索与回溯法)是一种选优搜索法,按选优条件向前搜索,以达到目标.但当探索到某一步时,发现原先选择并不优或达不到目标,就退回一步又一次选择,这样的走不通就退回再走的技术为回溯法,而满足回溯条件的某个状态的点称为"回溯点". 在包括问题的全部解的解空间树中,依照深度优先搜索的策略,从根结点出发深度探索解空间树.当探索到某一结点时,要先推断该结点是否包括问题的解,假设包括,就从该结点出发继续探索下去,假设该结点不包括问题的解,则逐层向其祖先结点回溯.(事实上回溯法就…

题目: Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = 3, a solution set is: "((()))", "(()())", "(())()", "()(())", "()()()" 题解:…

题目: Given two binary strings, return their sum (also a binary string). For example, a = "11" b = "1" Return "100". 题解: 二进制加法都是从最低位(从右加到左).所以对两个字符串要从最后一位开始加,如果遇见长度不一的情况,就把短的字符串高位补0. 每轮计算要加上进位,最后跳出循环后要坚持进位是否为1,以便更新结果. 代码如下(from…

题目: Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring. For "(()", the longest valid parentheses substring is "()", which has length = 2. Another example…