[LeetCode] 241. Different Ways to Add Parentheses

轉自LeetCode

Given a string expression of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. You may return the answer in any order.

 

Example 1:

Input: expression = "2-1-1"
Output: [0,2]
Explanation:
((2-1)-1) = 0 
(2-(1-1)) = 2

Example 2:

Input: expression = "2*3-4*5"
Output: [-34,-14,-10,-10,10]
Explanation:
(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

 

Constraints:

• 1 <= expression.length <= 20
• expression consists of digits and the operator '+', '-', and '*'.
• All the integer values in the input expression are in the range [0, 99].

Solution

這題也能看成是 95. Unique Binary Search Trees II 的衍生題

這題用 recursion 的方式來解會比較容易理解

能放入括號的，一定是運算符號的左右兩側，也就是 (...) + (...) or (...) - (...) or (...) * (...)

而括號裡面，也可以再用這種方式再去切分

所以可以想成是 binary tree 的資料結構

每當遇到運算符號，就去切分成左右兩段，找到計算的值後，再一個merge回來

kotlin(參考解答)

	class Solution {
	private val map = mutableMapOf<String,List<Int>>()
	fun diffWaysToCompute(expression: String): List<Int> {
	if (map[expression] != null) {
	return map[expression]!!
	}
	val ans = mutableListOf<Int>()
	for(i in expression.indices) {
	if(expression[i] == '+' || expression[i] == '-' || expression[i] == '*') {
	val left = diffWaysToCompute(expression.substring(0,i))
	val right = diffWaysToCompute(expression.substring(i+1))
	for(j in left.indices) {
	for(k in right.indices) {
	when {
	expression[i] == '+' -> ans.add(left[j] + right[k])
	expression[i] == '-' -> ans.add(left[j] - right[k])
	expression[i] == '*' -> ans.add(left[j] * right[k])
	}
	}
	}
	}
	}
	if(ans.isEmpty()) {
	ans.add(expression.toInt())
	}
	map[expression] = ans
	return ans
	}
	}

view raw different_ways_to_add_parentheses.kt hosted with ❤ by GitHub