1307. Verbal Arithmetic Puzzle

问题：

给定一组字符串，和一个结果字符串，使用0～9对字母进行编码。

使得字符串数组相加后，结果=结果字符串。

求是否可能存在这样的编码。

• Each character is decoded as one digit (0 - 9).
• Every pair of different characters they must map to different digits.
• Each words[i] and result are decoded as one number without leading zeros.（二位数以上的数，不得以0开头）
• Sum of numbers on left side (words) will equal to the number on right side (result). 

Example 1:
Input: words = ["SEND","MORE"], result = "MONEY"
Output: true
Explanation: Map 'S'-> 9, 'E'->5, 'N'->6, 'D'->7, 'M'->1, 'O'->0, 'R'->8, 'Y'->'2'
Such that: "SEND" + "MORE" = "MONEY" ,  9567 + 1085 = 10652

Example 2:
Input: words = ["SIX","SEVEN","SEVEN"], result = "TWENTY"
Output: true
Explanation: Map 'S'-> 6, 'I'->5, 'X'->0, 'E'->8, 'V'->7, 'N'->2, 'T'->1, 'W'->'3', 'Y'->4
Such that: "SIX" + "SEVEN" + "SEVEN" = "TWENTY" ,  650 + 68782 + 68782 = 138214

Example 3:
Input: words = ["THIS","IS","TOO"], result = "FUNNY"
Output: true

Example 4:
Input: words = ["LEET","CODE"], result = "POINT"
Output: false
 
Constraints:
2 <= words.length <= 5
1 <= words[i].length, result.length <= 7
words[i], result contain only uppercase English letters.
The number of different characters used in the expression is at most 10.

　　

解法：Backtracking（回溯算法）

• 状态：每个字符串的第pos位，相加后+进位flg后，得到的ans是否=result[pos]
• 选择：
  • 加数：w[pos]：未使用过的所有数digopt[i]==1（可用，0为不可用）
    • 递归，继续找本 pos 位。
  • 结果：result[pos]：w1[pos]+..+wn[pos]+flg = ans? && digopt[ans]==1 ：将ans加入编码中。
    • 递归，找下一位 pos+1 位
• 递归结束条件：pos=result的最高位。
  • 若 flg==0 则找到一个解。

⚠️ 注意：

除去两位以上且第一位为0的所有编码可能。

数位加法和字符串位置相反，因此pos代表数位（0:个位，1:十位...） spos代表字符串位（0:最高位）

spos = w.length()-1-pos;

spos<0时，说明加数位数已经不够，相加时，跳过该数。

代码参考：

class Solution {
public:
    //int count = 0;
    /*void print_intend() {
        for(int i=0; i<count; i++) {
            printf("  ");
        }
    }*/
    void backtrack(bool& res, int pos, int flg, vector<int>& map, vector<int>& optdig,
              vector<string>& words, string& result) {
        if(pos == result.size()) {
            if(flg==0){
                if(result.length() != 1 && map[result[0]-'A'] == 0) res = false;
                else res = true;
            } 
          //  print_intend();
          //  printf("return res=%d
",res);
            return;
        }
        int ans = flg;
        int spos;
        for(string w:words) {
            spos = w.length()-1-pos;
            if(spos<0) continue;
            if(map[w[spos]-'A'] != -1) ans+=map[w[spos]-'A'];
            else {
                for(int i=0; i<10; i++) {
                    if(w[spos]==w[0] && i==0 && w.length() != 1) continue;
                    if(optdig[i]==0) continue;
                    map[w[spos]-'A'] = i;
                    optdig[i] = 0;//used, unavaliable
            //count++;
            //print_intend();
            //printf("[%c]=%d
",w[spos],i);
            //print_intend();
            //printf("pos=%d, optdig.size=%d, flg=%d, [line:%d]
",pos,optdig.size(),flg, __LINE__);
                    backtrack(res, pos, flg, map, optdig, words, result);
            //count--;
                    optdig[i] = 1;//can use, avaliable
                    map[w[spos]-'A'] = -1;
                    if(res==true) return;
                }
            //print_intend();
            //printf("return res=%d [try param final failure]
",res);
                return;
            }
        }
        flg = ans/10;
        ans = ans%10;
        spos = result.length()-1-pos;
        if(map[result[spos]-'A'] == ans) {
            //count++;
            //print_intend();
            //printf("pos=%d, optdig.size=%d, flg=%d, [line:%d]
",pos,optdig.size(),flg, __LINE__);
            backtrack(res, pos+1, flg, map, optdig, words, result);
           // count--;
        } else if(map[result[spos]-'A'] != -1 || optdig[ans] == 0) {
            //print_intend();
            //printf("return res=%d [try failure]
",res);
            return;
        } else if(result[spos]==result[0] && ans==0 && result.length() != 1) {
            //print_intend();
            //printf("return res=%d [try head 0 failure]
",res);
            return;
        } else {
            map[result[spos]-'A'] = ans;
            optdig[ans] = 0;//used, unavaliable
            //count++;
            //print_intend();
            //printf("[%c]=%d
",result[spos],ans);
            //print_intend();
            //printf("pos=%d, optdig.size=%d, flg=%d, [line:%d]
",pos,optdig.size(),flg, __LINE__);
                backtrack(res, pos+1, flg, map, optdig, words, result);
            //count--;
                optdig[ans] = 1;//can use, avaliable
                map[result[spos]-'A'] = -1;
                if(res==true) return;
        }
            //print_intend();
            //printf("return res=%d [try res final failure]
",res);
        return;
    }
    bool isSolvable(vector<string>& words, string result) {
        bool res = false;
        vector<int> map(26,-1);
        vector<int> optdig(10,1);
        for(string w:words) {
            if(w.length()>result.length()) return false;
        }
        backtrack(res, 0, 0, map, optdig, words, result);
        return res;
    }
};