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: 
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]

解法类似 95题：http://www.jianshu.com/p/948c40b3444d

Solution1：Divide and Conquer

从每个operator将input_str分成左右两部分part1, part2 (Divide)，recursiely计算part1和part2的组合结果，再将这两个组合结果Conquer构成从当前operator分的结果。对于每个operator都这样遍历一遍。
Time Complexity: ？ Space Complexity: ？

Solution2：Divide and Conquer + Hashmap

Solution1中含有重复计算，2在Solution1的基础上 将已求过的String的结果List Mapping保存下来，缓存避免重复计算。
Time Complexity: ？ Space Complexity: ？

Solution3：DP写法

dp[i][j] stores all possible results from the i-th integer to the j-th integer (inclusive) in the list

                           /   length:d     \
                           [i->       j->    ]
操作数数组: [a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a..]

更新方式：遍历从d=0..N(不同长度的子串)，对于每个d，two loop遍历d范围中的i, j，求出dp[i][i+d]: dp[i][i+d].add(leftNum+rightNum); leftNum is from dp[i][j] list, and rightNum is from dp[j+1][i+d] list.
reference: https://discuss.leetcode.com/topic/26076/java-recursive-9ms-and-dp-4ms-solution
Time Complexity: ？ Space Complexity: ？

Solution1 Code：

class Solution {
    public List diffWaysToCompute(String input) {
        List result = new LinkedList();
        for(int i = 0; i < input.length(); i++) {
            char c = input.charAt(i);
            if(c == '+' || c == '-' || c == '*') {
                String part1 = input.substring(0, i);
                String part2 = input.substring(i + 1);
                List part1_ret = diffWaysToCompute(part1);
                List part2_ret = diffWaysToCompute(part2);
                for(Integer p1: part1_ret) {
                    for(Integer p2: part2_ret) {
                        int r = 0;
                        switch(c) {
                            case '+':
                                r = p1 + p2;
                                break;
                            case '-':
                                r = p1 - p2;
                                break;
                            case '*':
                                r = p1 * p2;
                                break;
                        }
                        result.add(r);
                    }
                }
            }
        }
        // if there is only one operand, no operator
        if(result.size() == 0) {
            result.add(Integer.valueOf(input));
        }
        return result;
    }
}

Solution2 Code：

public class Solution {
    Map> map;
    
    public List diffWaysToCompute(String input) {
        map = new HashMap<>();
        List result = new ArrayList<>();
        for (int i = 0; i < input.length(); i++) {
            char c = input.charAt(i);
            if (c == '+' || c == '-' || c == '*') {
                String part1 = input.substring(0, i);
                String part2 = input.substring(i + 1);
                List part1_ret = map.getOrDefault(part1, diffWaysToCompute(p1));
                List part2_ret = map.getOrDefault(part2, diffWaysToCompute(p2));
                for (Integer p1 : part1_ret) {
                    for (Integer p2 : part2_ret) {
                        int r = 0;
                        switch (c) {
                            case '+':
                                r = p1 + p2;
                                break;
                            case '-':
                                r = p1 - p2;
                                break;
                            case '*':
                                r = p1 * i2;
                                break;
                        }
                        res.add(r);
                    }
                }
            }
        }
        if (res.size() == 0) {
            res.add(Integer.valueOf(input));
        }
        map.put(input, result);
        return result;
    }
}

Solution3 Code：

public List diffWaysToCompute(String input) {
    List result=new ArrayList<>();
    if(input==null||input.length()==0)  return result;
    List ops=new ArrayList<>();
    for(int i=0; i[][] dp=(ArrayList[][]) new ArrayList[N][N];
    for(int d=0; d();
                dp[i][i].add(Integer.valueOf(ops.get(i*2)));
            }
            continue;
        }
        for(int i=0; i();
            for(int j=i; j left=dp[i][j], right=dp[j+1][i+d];
                String operator=ops.get(j*2+1);
                for(int leftNum:left)
                    for(int rightNum:right){
                        if(operator.equals("+"))
                            dp[i][i+d].add(leftNum+rightNum);
                        else if(operator.equals("-"))
                            dp[i][i+d].add(leftNum-rightNum);
                        else
                            dp[i][i+d].add(leftNum*rightNum);
                    }
            }
        }
    }
    return dp[0][N-1];
}