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, 0        while index < len(input) and input[index].isdigit():            val *= 10            val += int(input[index])            index += 1        if index == len(input):            result.append(val)            return result                    for 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 result            def compute(self, a, b, op):        if op == '+':            return a + b        elif op == '-':            return a - b        elif op == '*':            return a * b        else:            return 1

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