Different Ways to Add Parentheses(leetcode)

Different Ways to Add Parentheses

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题目

leetcode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]

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解决

想法：这道题采用分治算法，以符号+、-、*为标志位，当遇到这三个符号时候，以符号为分界线，将式子分为左右两部分，分别计算结果，之后再合并，得到最终结果。具体思路可以参照上面题目例子给出的过程

class Solution {public:    vector<int> diffWaysToCompute(string input) {        int len = input.size();        vector<int> result;        for (int k = 0; k < len; k++) {            if (input[k] == '+' || input[k] == '-' || input[k] == '*') { // 分割式子                vector<int> left = diffWaysToCompute(input.substr(0, k)); // 左边式子的结果                vector<int> right = diffWaysToCompute(input.substr(k + 1)); // 右边式子的结果                int n1 = left.size();                int n2 = right.size();                for (int i = 0; i < n1; i++) {                    for (int j = 0; j < n2; j++) {                        if (input[k] == '+') {                            result.push_back(left[i] + right[j]);                        } else if (input[k] == '-') {                            result.push_back(left[i] - right[j]);                        } else if (input[k] == '*') {                            result.push_back(left[i] * right[j]);                        }                    }                }            }        }        if (result.empty()) { // 输入的字符串没有符号，只有数字            result.push_back(atoi(input.c_str())); // 将字符串变成数字        }        return result;    }};