LeetCode-241. Different Ways to Add Parentheseshttps://leetcode.com/problems/different-ways-to-add-parentheses/

題目描述

Given a string?expression?of numbers and operators, return?all possible results from computing all the different possible ways to group numbers and operators. You may return the answer in?any order.

Example 1:

Input: expression = "2-1-1" Output: [0,2] Explanation: ((2-1)-1) = 0 (2-(1-1)) = 2

Example 2:

Input: expression = "2*3-4*5" Output: [-34,-14,-10,-10,10] Explanation: (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

Constraints:

• 1 <= expression.length <= 20
• expression?consists of digits and the operator?'+',?'-', and?'*'.
• All the integer values in the input expression are in the range?[0, 99].

解題思路

【C++】

1. 分治遞歸

class Solution {unordered_map<string, vector<int>> m; public:vector<int> diffWaysToCompute(string expression) {if (m.count(expression)) return m[expression];vector<int> ways;for (int i = 0; i < expression.length(); ++i) {char c = expression[i];if (c == '+' || c == '-' || c == '*') {vector<int> left = diffWaysToCompute(expression.substr(0, i));vector<int> right = diffWaysToCompute(expression.substr(i + 1));for (const int & l : left) {for (const int & r : right) {switch (c) {case '+': ways.push_back(l + r); break;case '-': ways.push_back(l - r); break; case '*': ways.push_back(l * r); break; }}}}}if (ways.empty()) {ways.push_back(stoi(expression));}m[expression] = ways;return ways;} };

2. 動態規劃

class Solution { public:vector<int> diffWaysToCompute(string expression) {vector<int> data;vector<char> ops;int num = 0;char op = ' ';istringstream ss(expression + "+");while (ss >> num && ss >> op) {data.push_back(num);ops.push_back(op);}int n = data.size();vector<vector<vector<int>>> dp(n, vector<vector<int>>(n, vector<int>()));for (int i = 0; i < n; ++i) {dp[i][i].push_back(data[i]);for (int j = i - 1; j >= 0; --j) {for (int k = j; k < i; k += 1) {for (auto left : dp[j][k]) {for (auto right : dp[k+1][i]) {int val = 0;switch (ops[k]) {case '+': val = left + right; break;case '-': val = left - right; break;case '*': val = left * right; break;}dp[j][i].push_back(val);}}}}}return dp[0][n-1];} };

【Java】

class Solution {private Map<String, List<Integer>> m = new ConcurrentHashMap<>();public List<Integer> diffWaysToCompute(String expression) {if (m.containsKey(expression)) {return m.get(expression);}List<Integer> ways = new ArrayList<>();for (int i = 0; i < expression.length(); ++i) {char c = expression.charAt(i);if (c == '+' || c == '-' || c == '*') {List<Integer> left = diffWaysToCompute(expression.substring(0, i));List<Integer> right = diffWaysToCompute(expression.substring(i + 1));for (int l : left) {for (int r : right) {switch (c) {case '+': ways.add(l + r); break;case '-': ways.add(l - r); break; case '*': ways.add(l * r); break; }}}}}if (ways.isEmpty()) {ways.add(Integer.parseInt(expression));}m.put(expression, ways);return ways;} }

總結

以上是生活随笔為你收集整理的LeetCode-241. Different Ways to Add Parentheses [C++][Java]的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。