leetcode 241. Different Ways to Add Parentheses

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]

这个问题就用递归来解决啦。

public List<Integer> diffWaysToCompute(String input) {    List<Integer> res = new ArrayList<Integer>();    for (int i = 0; i < input.length(); i++) {        char c = input.charAt(i);        if (c == '-' || c == '+' || c == '*') {            String a = input.substring(0, i);            String b = input.substring(i + 1);            List<Integer> al = diffWaysToCompute(a);            List<Integer> bl = diffWaysToCompute(b);            for (int x : al) {                for (int y : bl) {                    if (c == '-') {                        res.add(x - y);                    } else if (c == '+') {                        res.add(x + y);                    } else if (c == '*') {                        res.add(x * y);                    }                }            }        }    }    if (res.size() == 0) {  //如果input是一个单独的数，没有符号    res.add(Integer.valueOf(input));    }    return res;}

大神不仅用递归，而且还增加了记忆，避免递归时重复计算同一个字符串的result list.

public class Solution {        Map<String, List<Integer>> map = new HashMap<>();        public List<Integer> diffWaysToCompute(String input) {        List<Integer> res = new ArrayList<>();        for (int i = 0; i < input.length(); i++) {            char c = input.charAt(i);            if (c == '+' || c == '-' || c == '*') {                String p1 = input.substring(0, i);                String p2 = input.substring(i + 1);                List<Integer> l1 = map.getOrDefault(p1, diffWaysToCompute(p1));                List<Integer> l2 = map.getOrDefault(p2, diffWaysToCompute(p2));                for (Integer i1 : l1) {                    for (Integer i2 : l2) {                        int r = 0;                        switch (c) {                            case '+':                                r = i1 + i2;                                break;                            case '-':                                r = i1 - i2;                                break;                            case '*':                                r = i1 * i2;                                break;                        }                        res.add(r);                    }                }            }        }        if (res.size() == 0) {            res.add(Integer.valueOf(input));        }        map.put(input, res);        return res;    }}