282. Expression Add Operators¶
Difficulty: Hard
LeetCode Problem View on GitHub
282. Expression Add Operators
Hard
Given a string num that contains only digits and an integer target, return all possibilities to insert the binary operators '+', '-', and/or '*' between the digits of num so that the resultant expression evaluates to the target value.
Note that operands in the returned expressions should not contain leading zeros.
Example 1:
Input: num = "123", target = 6 Output: ["1*2*3","1+2+3"] Explanation: Both "1*2*3" and "1+2+3" evaluate to 6.
Example 2:
Input: num = "232", target = 8 Output: ["2*3+2","2+3*2"] Explanation: Both "2*3+2" and "2+3*2" evaluate to 8.
Example 3:
Input: num = "3456237490", target = 9191 Output: [] Explanation: There are no expressions that can be created from "3456237490" to evaluate to 9191.
Constraints:
1 <= num.length <= 10numconsists of only digits.-231 <= target <= 231 - 1
Solution¶
public class Solution {
public List<String> addOperators(String num, int target) {
List<String> rst = new ArrayList<String>();
if(num == null || num.length() == 0) return rst;
helper(rst, "", num, target, 0, 0, 0);
return rst;
}
public void helper(List<String> rst, String path, String num, int target, int pos, long eval, long multed){
if(pos == num.length()){
if(target == eval) rst.add(path);
return;
}
for(int i = pos; i < num.length(); i++){
if(i != pos && num.charAt(pos) == '0') break;
long cur = Long.parseLong(num.substring(pos, i + 1));
if(pos == 0) helper(rst, path + cur, num, target, i + 1, cur, cur);
else{
helper(rst, path + "+" + cur, num, target, i + 1, eval + cur , cur);
helper(rst, path + "-" + cur, num, target, i + 1, eval -cur, -cur);
helper(rst, path + "*" + cur, num, target, i + 1, eval - multed + multed * cur, multed * cur );
}
}
}
}
Complexity Analysis¶
- Time Complexity:
O(?) - Space Complexity:
O(?)
Approach¶
Detailed explanation of the approach will be added here