[LeetCode] 241. Different Ways to Add Parentheses 添加括号的不同方式
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]
给定一个含有数字和运算符的表达式，运算符可以是加减乘，在任意位置添加括号，求出所有可能的表达式值。
解法：递归
Java:
public class Solution {
public List<Integer> diffWaysToCompute(String input) {
List<Integer> result = new ArrayList<>();
if (input == null || input.length() == 0) {
return result;
}
for (int i = 0; i < input.length(); i++) {
char c = input.charAt(i);
if (!isOperator(c)) {
continue;
}
List<Integer> left = diffWaysToCompute(input.substring(0, i));
List<Integer> right = diffWaysToCompute(input.substring(i + 1));
for (int num1 : left) {
for (int num2 : right) {
int val = calculate(num1, num2, c);
result.add(val);
}
}
}
// only contains one number
if (result.isEmpty()) {
result.add(Integer.parseInt(input));
}
return result;
}
private int calculate(int num1, int num2, char operator) {
int result = 0;
switch(operator) {
case '+' : result = num1 + num2;
break;
case '-' : result = num1 - num2;
break;
case '*' : result = num1 * num2;
break;
}
return result;
}
private boolean isOperator(char operator) {
return (operator == '+') || (operator == '-') || (operator == '*');
}
}
Python:
import operator
import re
class Solution:
# @param {string} input
# @return {integer[]}
def diffWaysToCompute(self, input):
tokens = re.split('(\D)', input)
nums = map(int, tokens[::2])
ops = map({'+': operator.add, '-': operator.sub, '*': operator.mul}.get, tokens[1::2])
lookup = [[None for _ in xrange(len(nums))] for _ in xrange(len(nums))]
def diffWaysToComputeRecu(left, right):
if left == right:
return [nums[left]]
if lookup[left][right]:
return lookup[left][right]
lookup[left][right] = [ops[i](x, y)
for i in xrange(left, right)
for x in diffWaysToComputeRecu(left, i)
for y in diffWaysToComputeRecu(i + 1, right)]
return lookup[left][right]
return diffWaysToComputeRecu(0, len(nums) - 1)
Python:
class Solution:
# @param {string} input
# @return {integer[]}
def diffWaysToCompute(self, input):
lookup = [[None for _ in xrange(len(input) + 1)] for _ in xrange(len(input) + 1)]
ops = {'+': operator.add, '-': operator.sub, '*': operator.mul}
def diffWaysToComputeRecu(left, right):
if lookup[left][right]:
return lookup[left][right]
result = []
for i in xrange(left, right):
if input[i] in ops:
for x in diffWaysToComputeRecu(left, i):
for y in diffWaysToComputeRecu(i + 1, right):
result.append(ops[input[i]](x, y))
if not result:
result = [int(input[left:right])]
lookup[left][right] = result
return lookup[left][right]
return diffWaysToComputeRecu(0, len(input))
C++:
class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> res;
for (int i = 0; i < input.size(); ++i) {
if (input[i] == '+' || input[i] == '-' || input[i] == '*') {
vector<int> left = diffWaysToCompute(input.substr(0, i));
vector<int> right = diffWaysToCompute(input.substr(i + 1));
for (int j = 0; j < left.size(); ++j) {
for (int k = 0; k < right.size(); ++k) {
if (input[i] == '+') res.push_back(left[j] + right[k]);
else if (input[i] == '-') res.push_back(left[j] - right[k]);
else res.push_back(left[j] * right[k]);
}
}
}
}
if (res.empty()) res.push_back(atoi(input.c_str()));
return res;
}
};
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