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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