282. Expression Add Operators (Hard)

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 -> []

Solution: Divide and Conquer

这道题给了我们一个只由数字组成的字符串，让我们再其中添加+,-或*号来形成一个表达式，该表达式的计算和为给定了target值，让我们找出所有符合要求的表达式来。题目中给的几个例子其实并不好，很容易让人误以为是必须拆成个位数字，其实不是的，比如"123", 15能返回"12+3"，说明连着的数字也可以。如果非要在过往的题中找一道相似的题，我觉得跟Combination Sum II 组合之和之二很类似。不过这道题要更复杂麻烦一些。还是用递归来解题，我们需要两个变量diff和curNum，一个用来记录将要变化的值，另一个是当前运算后的值，而且它们都需要用long long型的，因为字符串转为int型很容易溢出，所以我们用长整型。对于加和减，diff就是即将要加上的数和即将要减去的数的负值，而对于乘来说稍有些复杂，此时的diff应该是上一次的变化的diff乘以即将要乘上的数，有点不好理解，那我们来举个例子，比如2+3*2，即将要运算到乘以2的时候，上次循环的curNum = 5, diff = 3, 而如果我们要算这个乘2的时候，新的变化值diff应为3*2=6，而我们要把之前+3操作的结果去掉，再加上新的diff，即(5-3)+6=8，即为新表达式2+3*2的值，有点难理解，大家自己一步一步推算吧。

还有一点需要注意的是，如果输入为"000",0的话，容易出现以下的错误：

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"]

我们可以看到错误的结果中有0开头的字符串出现，明显这不是数字，所以我们要去掉这些情况，过滤方法也很简单，我们只要判断长度大于1且首字符是‘0’的字符串，将其滤去即可，参见代码如下：

verson1: 306ms (slow)

see next version

class Solution {
public:
    vector<string> addOperators(string num, int target) {
        vector<string> res;
        dfs(num, target, 0, 0, "", res);
        return res;
    }

    void dfs(string num, int target, long diff, 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; // remove numbers like "000"
            string next = num.substr(i);
            if (out.size() > 0) {
                // +
                dfs(next, target, stol(cur), curNum+stol(cur), out+"+"+cur, res);
                // -
                dfs(next, target, -stol(cur), curNum-stol(cur), out+"-"+cur, res);
                // *
                dfs(next, target, diff*stol(cur), curNum-diff+diff*stol(cur), out+"*"+cur, res);
            } else {
                dfs(next, target, stol(cur), stol(cur), cur, res);
            }
        }
    }
};

version2: 126ms

use int pos, no need to pass next string

class Solution {
    vector<string> ans;

    void dfs(string& num, int target, int pos, long cur, long prev, string out, vector<string>& res) {
        if (pos == num.size()) {
            if (cur == target) res.push_back(out);
            return;
        }
        for (int i = pos; i < num.size(); ++i) {
            if (i > pos && num[pos] == '0') return;
            string tmp = num.substr(pos, i-pos+1);
            long val = stol(tmp);
            if (pos != 0) {
                dfs(num, target, i+1, cur+val, val, out+'+'+tmp, res);
                dfs(num, target, i+1, cur-val, -val, out+'-'+tmp, res);
                dfs(num, target, i+1, cur-prev+prev*val, prev*val, out+'*'+tmp, res);
            } else {
                dfs(num, target, i+1, val, val, tmp, res);
            }
        }

    }
public:
    vector<string> addOperators(string num, int target) {
        vector<string> res;
        dfs(num, target, 0, 0, 0, "", res);
        return res;
    }
};