LeetCode 241 Different Ways to Add Parentheses

题目描述

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]

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

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分析

使用递归，对于每个运算符号+ - *，将string分成两部分（不包括这个运算符号）。分别计算出两部分的结果，再运用这个运算符号进行运算。

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代码

    public List<Integer> diffWaysToCompute(String input) {        List<Integer> rt = new LinkedList<Integer>();        int len = input.length();        for (int i = 0; i < len; i++) {            if (input.charAt(i) == '-' || input.charAt(i) == '*'                    || input.charAt(i) == '+') {                String part1 = input.substring(0, i);                String part2 = input.substring(i + 1);                List<Integer> part1Ret = diffWaysToCompute(part1);                List<Integer> part2Ret = diffWaysToCompute(part2);                for (Integer p1 : part1Ret) {                    for (Integer p2 : part2Ret) {                        int c = 0;                        switch (input.charAt(i)) {                        case '+':                            c = p1 + p2;                            break;                        case '-':                            c = p1 - p2;                            break;                        case '*':                            c = p1 * p2;                        }                        rt.add(c);                    }                }            }        }        if (rt.size() == 0) {            rt.add(Integer.valueOf(input));        }        return rt;    }