LeetCode Different Ways to Add Parentheses

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]

题意

输入一个只有数字、操作符(+ - *)字符串，通过任意添加括号，求出这个表达式的所有可能值

思路

直接暴力计算

对于当前一个字符串，枚举最后计算的那个操作符。

比如12*34+56-78+90

分别枚举最后计算的操作符为

*

+

-

+

然后计算出当前操作符左边、右边字符串的所有可能值，然后合并左右就是当前问题的解

ps markdown真是太好用了

代码

class Solution {
public:
vector<int> diffWaysToCompute(string input) {
_expression = input;
return dfs(0, input.length()-1);
}

private :
string _expression;
vector<int> dfs(int left, int right) {
vector<int> result;
bool hasOperators = false;
for (int index = left+1; index < right; index ++) {
char value = _expression[index];
if (value == '*' || value == '+' || value == '-') {
hasOperators = true;
break;
}
}

if (!hasOperators) {
int value = 0;
for (int index = left; index <= right; index ++) {
assert(_expression[index] >= '0' && _expression[index] <= '9');
value = value * 10 + _expression[index] - '0';
}
result.push_back(value);
return result;
}

for (int index = left + 1; index < right; index ++) {
char value = _expression[index];
if (value == '*' || value == '+' || value == '-') {
vector<int> leftValues = dfs(left, index-1);
vector<int> rightValues = dfs(index+1, right);
for (vector<int>::iterator pleft = leftValues.begin(); pleft != leftValues.end(); pleft ++) {
for (vector<int>::iterator pright = rightValues.begin(); pright != rightValues.end(); pright ++) {
int ans = 0;
if (value == '*') {
ans = *pleft * *pright;
} else if (value == '+') {
ans = *pleft + *pright;
} else {
ans = *pleft - *pright;
}
result.push_back(ans);
}
}
}
}

return result;
}
};