282. Expression Add Operators - jiejackyzhang/leetcode-note GitHub Wiki

Given a string that contains only digits 0-9 and a target value, return all possibilities to add binary operators (not unary) +, -, or * between the digits so they evaluate to the target value.

Examples:

"123", 6 -> ["1+2+3", "1*2*3"] 
"232", 8 -> ["2*3+2", "2+3*2"]
"105", 5 -> ["1*0+5","10-5"]
"00", 0 -> ["0+0", "0-0", "0*0"]
"3456237490", 9191 -> []

解题思路为backtracking。

需要注意这几点：

  1. overflow：采用long来避免溢出。
  2. 处理连续的0。
  3. 必须保存下一次乘法递归要用的乘数。
public class Solution {
    public List<String> addOperators(String num, int target) {
        List<String> res = new ArrayList<>();
        if(num == null || num.length() == 0) return res;
        helper(num, target, 0, 0, 0, res, "");
        return res;
    }
    
    private void helper(String num, int target, int pos, long eval, long multed, List<String> res, String path) {
        if(pos == num.length()) {
            if(eval == target) {
                res.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(num, target, i+1, cur, cur, res, path+cur);
            } else {
                helper(num, target, i+1, eval+cur, cur, res, path+"+"+cur);
                helper(num, target, i+1, eval-cur, -cur, res, path+"-"+cur);
                helper(num, target, i+1, eval-multed+multed*cur, multed*cur, res, path+"*"+cur);
            }
        }
    }
}
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