LeetCode 题解（150）: Different Ways to Add Parentheses

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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

Output: [0, 2]

Example 2

Input: "2*3-4*5"

(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

Output: [-34, -14, -10, -10, 10]

题解：

网上学来的做法，左右递归，递归终止条件是当前string只包含一个数，不包含符号时，直接返回该数。

C++版：

class Solution {
public:vector<int> diffWaysToCompute(string input) {vector<int> result;if (input.length() == 0)return result;int val = 0, index = 0;while(index < input.length() && isdigit(input[index])) {val *= 10;val += input[index++] - '0';}if(index == input.length()) {result.push_back(val);return result;}for(int i = 0; i < input.length(); i++) {if(!isdigit(input[i])) {vector<int> left = diffWaysToCompute(input.substr(0, i));vector<int> right = diffWaysToCompute(input.substr(i+1, input.length() - i - 1));for(int j = 0; j < left.size(); j++) {for(int k = 0; k < right.size(); k++) {result.push_back(compute(left[j], right[k], input[i]));}}}}return result;}int compute(int a, int b, char op) {switch (op) {case '+': return a + b;case '-': return a - b;case '*': return a * b;default: return 1;}}
};

Java版：

public class Solution {public List<Integer> diffWaysToCompute(String input) {List<Integer> result = new ArrayList<Integer>();int val = 0, index = 0;while(index < input.length() && Character.isDigit(input.charAt(index))) {val *= 10;val += input.charAt(index++) - '0';}if(index == input.length()) {result.add(val);return result;}for(int i = 0; i < input.length(); i++) {if(!Character.isDigit(input.charAt(i))) {List<Integer> left = diffWaysToCompute(input.substring(0, i));List<Integer> right = diffWaysToCompute(input.substring(i+1, input.length()));for(int j = 0; j < left.size(); j++) {for(int k = 0; k < right.size(); k++) {result.add(compute(left.get(j), right.get(k), input.charAt(i)));}}}}return result;}int compute(int a, int b, char op) {switch(op) {case '+': return a + b;case '-': return a - b;case '*': return a * b;default: return 1;}}
}

Python版：

class Solution:# @param {string} input# @return {integer[]}def diffWaysToCompute(self, input):result = []val, index = 0, 0while index < len(input) and input[index].isdigit():val *= 10val += int(input[index])index += 1if index == len(input):result.append(val)return resultfor i in range(len(input)):if not input[i].isdigit():left = self.diffWaysToCompute(input[:i])right = self.diffWaysToCompute(input[i+1:])for j in range(len(left)):for k in range(len(right)):result.append(self.compute(left[j], right[k], input[i]))return resultdef compute(self, a, b, op):if op == '+':return a + belif op == '-':return a - belif op == '*':return a * belse:return 1

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