Different Ways to Add Parentheses
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2022-05-13 20:07:42
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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]
输出所有可能的结果。我们采用分治法,当遇到字符‘+’ ,’-‘ ,’*‘ 时就把字符串分为两段,然后递归求出左半部分可右半部分能有的结果,然后在将左半部分和有半部分根据当前的字符求出当前可能的结果。当for循环结束就返回结果。代码如下:
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]
输出所有可能的结果。我们采用分治法,当遇到字符‘+’ ,’-‘ ,’*‘ 时就把字符串分为两段,然后递归求出左半部分可右半部分能有的结果,然后在将左半部分和有半部分根据当前的字符求出当前可能的结果。当for循环结束就返回结果。代码如下:
public class Solution {
public List<Integer> diffWaysToCompute(String input) {
List<Integer> list = new ArrayList<Integer>();
if(input == null || input.length() == 0) return list;
if(input.indexOf("+") < 0 && input.indexOf("-") < 0 && input.indexOf("*") < 0) {
list.add(Integer.parseInt(input));
return list;
}
for(int i = 0; i < input.length(); i++) {
char tem = input.charAt(i);
if(tem == '+' || tem == '-' || tem == '*') {
List<Integer> left = diffWaysToCompute(input.substring(0, i));
List<Integer> right = diffWaysToCompute(input.substring(i + 1, input.length()));
for(int m : left)
for(int n : right) {
switch(input.charAt(i)) {
case('+') :
list.add(m + n);
break;
case('-') :
list.add(m - n);
break;
case('*') :
list.add(m * n);
break;
}
}
}
}
return list;
}
}
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