[LeetCode] 282. Expression Add Operators 表达式增加操作符

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.

Example 1:

Input: num = "123", target = 6
Output: ["1+2+3", "1*2*3"] 

Example 2:

Input: num = "232", target = 8
Output: ["2*3+2", "2+3*2"]

Example 3:

Input: num = "105", target = 5
Output: ["1*0+5","10-5"]

Example 4:

Input: num = "00", target = 0
Output: ["0+0", "0-0", "0*0"]

Example 5:

Input: num = "3456237490", target = 9191
Output: []

给一个只由数字组成的字符串,在数字之间添加+,-或*号来组成一个表达式,使得该表达式的计算结果为给定了target值,找出所有符合要求的表达式。

解法:递归

This problem has a lot of edge cases to be considered:

overflow: we use a long type once it is larger than Integer.MAX_VALUE or minimum, we get over it.
0 sequence: because we can't have numbers with multiple digits started with zero, we have to deal with it too.
a little trick is that we should save the value that is to be multiplied in the next recursion.

Java:

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 );
            }
        }
    }
}  

Python:

class Solution(object):
    def addOperators(self, num, target):
        """
        :type num: str
        :type target: int
        :rtype: List[str]
        """
        result, expr = [], []
        val, i = 0, 0
        val_str = ""
        while i < len(num):
            val = val * 10 + ord(num[i]) - ord('0')
            val_str += num[i]
            # Avoid "00...".
            if str(val) != val_str:
                break
            expr.append(val_str)
            self.addOperatorsDFS(num, target, i + 1, 0, val, expr, result)
            expr.pop()
            i += 1
        return result

    def addOperatorsDFS(self, num, target, pos, operand1, operand2, expr, result):
        if pos == len(num) and operand1 + operand2 == target:
            result.append("".join(expr))
        else:
            val, i = 0, pos
            val_str = ""
            while i < len(num):
                val = val * 10 + ord(num[i]) - ord('0')
                val_str += num[i]
                # Avoid "00...".
                if str(val) != val_str:
                    break

                # Case '+':
                expr.append("+" + val_str)
                self.addOperatorsDFS(num, target, i + 1, operand1 + operand2, val, expr, result)
                expr.pop()

                # Case '-':
                expr.append("-" + val_str)
                self.addOperatorsDFS(num, target, i + 1, operand1 + operand2, -val, expr, result)
                expr.pop()

                # Case '*':
                expr.append("*" + val_str)
                self.addOperatorsDFS(num, target, i + 1, operand1, operand2 * val, expr, result)
                expr.pop()

                i += 1  

 C++:

class Solution {
public:
    vector<string> addOperators(string num, int target) {
        vector<string> res;
        addOperatorsDFS(num, target, 0, 0, "", res);
        return res;
    }
    void addOperatorsDFS(string num, int target, long long diff, long long curNum, string out, vector<string> &res) {
        if (num.size() == 0 && curNum == target) {
            res.push_back(out);
        }
        for (int i = 1; i <= num.size(); ++i) {
            string cur = num.substr(0, i);
            if (cur.size() > 1 && cur[0] == '0') return;
            string next = num.substr(i);
            if (out.size() > 0) {
                addOperatorsDFS(next, target, stoll(cur), curNum + stoll(cur), out + "+" + cur, res);
                addOperatorsDFS(next, target, -stoll(cur), curNum - stoll(cur), out + "-" + cur, res);
                addOperatorsDFS(next, target, diff * stoll(cur), (curNum - diff) + diff * stoll(cur), out + "*" + cur, res);
            } else {
                addOperatorsDFS(next, target, stoll(cur), stoll(cur), cur, res);
            }
        }
    }
};

  

 

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posted @ 2018-09-08 13:48  轻风舞动  阅读(782)  评论(0编辑  收藏  举报