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
class Solution { Map<String, List<Integer>> map = new HashMap<>(); public List<Integer> diffWaysToCompute(String input) { if(map.containsKey(input)) return map.get(input); List<Integer> ret = new ArrayList<>(); for (int i=0; i<input.length(); i++) { if (input.charAt(i) == '-' || input.charAt(i) == '*' || input.charAt(i) == '+' ) { String part1 = input.substring(0, i); String part2 = input.substring(i+1); List<Integer> part1Ret = diffWaysToCompute(part1); List<Integer> part2Ret = diffWaysToCompute(part2); for (Integer p1 : part1Ret) { for (Integer p2 : part2Ret) { int c = 0; switch (input.charAt(i)) { case '+': c = p1+p2; break; case '-': c = p1-p2; break; case '*': c = p1*p2; break; } ret.add(c); } } } } //意味着input此时是数字 if (ret.size() == 0) { ret.add(Integer.valueOf(input)); } //将ret和input添加到map中,如果有重复可以直接使用 map.put(input, ret); return ret; } }