共计 1461 个字符，预计需要花费 4 分钟才能阅读完成。
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”
Output: [0, 2]
Explanation:
((2-1)-1) = 0
(2-(1-1)) = 2
Example 2:
Input: “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
难度：medium
题目：给定一包含数字和操作符的字符串，计算并返回其所有数字与操作符组合的结果。有效的操作符为 +,-,*
思路：类似 unique binary tree, generate parentheses. 卡特兰数。
Runtime: 2 ms, faster than 84.52% of Java online submissions for Different Ways to Add Parentheses.Memory Usage: 38.6 MB, less than 18.35% of Java online submissions for Different Ways to Add Parentheses.
class Solution {
public List<Integer> diffWaysToCompute(String input) {
return diffWaysToCompute(input, 0, input.length());
}
private List<Integer> diffWaysToCompute(String str, int start, int end) {
if (!hasOperator(str, start, end)) {
List<Integer> result = new ArrayList<>();
result.add(Integer.parseInt(str.substring(start, end)));
return result;
}
List<Integer> root = new ArrayList<>();
for (int i = start; i < end; i++) {
char c = str.charAt(i);
if (‘+’ == c || ‘-‘ == c || ‘*’ == c) {
List<Integer> left = diffWaysToCompute(str, start, i);
List<Integer> right = diffWaysToCompute(str, i + 1, end);
for (Integer l: left) {
for (Integer r : right) {
if (‘+’ == c) {
root.add(l + r);
} else if (‘-‘ == c) {
root.add(l – r);
} else if (‘*’ == c) {
root.add(l * r);
}
}
}
}
}
return root;
}
private boolean hasOperator(String s, int start, int end) {
for (int i = start; i < end; i++) {
char c = s.charAt(i);
if (‘+’ == c || ‘-‘ == c || ‘*’ == c) {
return true;
}
}
return false;
}
}
