Leetcode282. Expression Add Operators

淡淡的烟草味﹌ 2022-06-08 08:21 201阅读 0赞

# Analysis #

Oveiously, in order to solve this question, we have to iterate through all of the possible cases at least once. DFS can be used to solve the iteration process.

The optimization we could made it to prune the unnecessary branches, e.g. there is not need to try 0 sequences longer than 1, because this is not a natural number.

And also, like always, we have to take into consideration the overflow problem for a integer-related calculation.

In addition, we have to also deal the the multiplication case differently since multiplicaiton operation has a higher priority than plus and minus. Thus if a multiplication shows up latter, we have to change the calculation sequence in our result. E.g. when we get “2+3” in the previous search and we want to check “2+3\*5” in the next step, we have to first subtract 3 from “2+3” and then add “3\*5” to the result.

## Time complexity: ##

Considering for every digit `0-9` in the number, we have 4 possible operator that can be assigned between them – `+`, `-`, ‘\*`and`join`, which means the previous digit and current digit will combined to become 1 number, e.g.`2 join 3`will be`23\`.

Based on the fact that there are `n` digits in total, there are `O(4^n)` possible cases that we need to check. Hence the time complexity of the program is `O(4^n)`.

# Implementation in Java: #

class Solution {
        public List<String> addOperators(String num, int target) {
            List<String> res = new ArrayList<>();
            if (num == null || num.length() == 0) {
                return res;
            }
    
            addOperators(res, "", num, target, 0, 0, 0);
            return res;
        }
    
        private void addOperators(List<String> res, String path, String num, int target, int start, long eval, long nextMulted) {
            if (start == num.length()) {
                if (eval == target) {
                    res.add(path);
                    return;
                }
            }
            for (int i = start; i < num.length(); i++) {
                if (i != start && num.charAt(start) == '0') {
                    break;
                }
                long curr = Long.parseLong(num.substring(start, i + 1));
                if (start == 0) {
                    addOperators(res, path + curr, num, target, i + 1, curr, curr);
                } else {
                    addOperators(res, path + "+" + curr, num, target, i + 1, eval + curr, curr);
                    addOperators(res, path + "-" + curr, num, target, i + 1, eval - curr, -curr);
                    addOperators(res, path + "*" + curr, num, target, i + 1, eval - nextMulted + nextMulted * curr, nextMulted * curr);
                }
            }
        }
    }