[LeetCode] Expression Add Operators 表達式增加操作符

 

Given a string that contains only digits 0-9 and a target value, return all possibilities to add binaryoperators (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: []

Credits:
Special thanks to @davidtan1890 for adding this problem and creating all test cases.

 

這道題給了我們一個只由數字組成的字符串，讓我們再其中添加+,-或*號來形成一個表達式，該表達式的計算和為給定了target值，讓我們找出所有符合要求的表達式來。看了題目中的例子1和2，很容易讓人誤以為是必須拆成個位數字，其實不是的，比如例子3中的 "105", 5能返回"10-5"，說明連着的數字也可以。如果非要在過往的題中找一道相似的題，我覺得跟 Combination Sum II 很類似。不過這道題要更復雜麻煩一些。還是用遞歸來解題，我們需要兩個變量diff和curNum，一個用來記錄將要變化的值，另一個是當前運算后的值，而且它們都需要用 long 型的，因為字符串轉為int型很容易溢出，所以我們用長整型。對於加和減，diff就是即將要加上的數和即將要減去的數的負值，而對於乘來說稍有些復雜，此時的diff應該是上一次的變化的diff乘以即將要乘上的數，有點不好理解，那我們來舉個例子，比如 2+3*2，即將要運算到乘以2的時候，上次循環的 curNum = 5, diff = 3, 而如果我們要算這個乘2的時候，新的變化值diff應為 3*2=6，而我們要把之前+3操作的結果去掉，再加上新的diff，即 (5-3)+6=8，即為新表達式 2+3*2 的值，有點難理解，大家自己一步一步推算吧。

還有一點需要注意的是，如果輸入為"000",0的話，容易出現以下的錯誤：

Wrong：["0+0+0","0+0-0","0+0*0","0-0+0","0-0-0","0-0*0","0*0+0","0*0-0","0*0*0","0+00","0-00","0*00","00+0","00-0","00*0","000"]

Correct：["0*0*0","0*0+0","0*0-0","0+0*0","0+0+0","0+0-0","0-0*0","0-0+0","0-0-0"]

我們可以看到錯誤的結果中有0開頭的字符串出現，明顯這不是數字，所以我們要去掉這些情況，過濾方法也很簡單，我們只要判斷長度大於1且首字符是‘0’的字符串，將其濾去即可，參見代碼如下：

 

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

 

類似題目：

Evaluate Reverse Polish Notation

Basic Calculator II

Basic Calculator

Different Ways to Add Parentheses

Target Sum

 

參考資料：

https://leetcode.com/problems/expression-add-operators/

https://leetcode.com/problems/expression-add-operators/discuss/71971/Accepted-C%2B%2B-Solution

https://leetcode.com/problems/expression-add-operators/discuss/71895/Java-Standard-Backtrace-AC-Solutoin-short-and-clear

 

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