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]
题目大意：
给定一个包含数字和运算符的字符串，返回所有对数字与运算符分组可能得到的计算结果。有效的运算符为 +, - 和 *
测试样例见题目描述。
解题思路：
解法I 分治法（Divide and conquer）
递归地将表达式按照运算符拆分为左右两半
类Python伪码：
def solve(input):
if input is digit:
return [input]
ans = [ ]
for x in input:
if x is operator:
left = solve(input[0 ~ x])
right = solve(input[x + 1 ~ len])
for l in left:
for r in right:
ans += [ calc(l, r, x) ]
return ans
Python代码：
class Solution(object):
def diffWaysToCompute(self, input):
"""
:type input: str
:rtype: List[int]
"""
if input.isdigit(): return [int(input)]
result = []
for i, c in enumerate(input):
if c in '+-*':
left = self.diffWaysToCompute(input[:i])
right = self.diffWaysToCompute(input[i+1:])
for l in left:
for r in right:
result.append(eval(str(l) + c + str(r)))
return result
上述解法可以通过记忆化进行加速。
Python代码：
class Solution(object):
def __init__(self):
self.cache = collections.defaultdict(list)
def diffWaysToCompute(self, input):
"""
:type input: str
:rtype: List[int]
"""
if input.isdigit(): return [int(input)]
if input in self.cache: return self.cache[input]
result = []
for i, c in enumerate(input):
if c in '+-*':
left = self.diffWaysToCompute(input[:i])
right = self.diffWaysToCompute(input[i+1:])
for l in left:
for r in right:
result.append(eval(str(l) + c + str(r)))
self.cache[input] = result
return result
解法II 回溯法（Backtracking），时间复杂度O(n!)，n为运算符的个数
通过dict保存有效的加括号方案，使用内置函数expr计算结果
Python代码：
class Solution:
# @param {string} input
# @return {integer[]}
def diffWaysToCompute(self, input):
exprDict = dict()
def dfs(nums, ops):
if ops:
for x in range(len(ops)):
dfs(nums[:x]+['('+nums[x]+ops[x]+nums[x+1]+')']+nums[x+2:],ops[:x]+ops[x+1:])
elif nums[0] not in exprDict:
exprDict[nums[0]] = eval(nums[0])
nums, ops = [], []
input = re.split(r'(\D)', input)
for x in input:
if x.isdigit():
nums.append(x)
else:
ops.append(x)
dfs(nums, ops)
return exprDict.values()
另附一段简洁的Python代码，学习一下：
摘自LeetCode Discuss：https://leetcode.com/discuss/48468/11-lines-python
Solution 1 - 48 ms
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])
def build(lo, hi):
if lo == hi:
return [nums[lo]]
return [ops[i](a, b)
for i in xrange(lo, hi)
for a in build(lo, i)
for b in build(i + 1, hi)]
return build(0, len(ops))
Solution 2 - 168 ms
class Solution:
# @param {string} input
# @return {integer[]}
def diffWaysToCompute(self, input):
return [eval(`a`+c+`b`)
for i, c in enumerate(input) if c in '+-*'
for a in self.diffWaysToCompute(input[:i])
for b in self.diffWaysToCompute(input[i+1:])] or [int(input)]
Solution 3 - 64 ms
class Solution:
# @param {string} input
# @return {integer[]}
def diffWaysToCompute(self, input):
return [a+b if c == '+' else a-b if c == '-' else a*b
for i, c in enumerate(input) if c in '+-*'
for a in self.diffWaysToCompute(input[:i])
for b in self.diffWaysToCompute(input[i+1:])] or [int(input)]
本文链接：http://bookshadow.com/weblog/2015/07/27/leetcode-different-ways-add-parentheses/
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