[LeetCode] Design Excel Sum Formula 设计Excel表格求和公式

Your task is to design the basic function of Excel and implement the function of sum formula. Specifically, you need to implement the following functions:

Excel(int H, char W): This is the constructor. The inputs represents the height and width of the Excel form. His a positive integer, range from 1 to 26. It represents the height. W is a character range from 'A' to 'Z'. It represents that the width is the number of characters from 'A' to W. The Excel form content is represented by a height * width 2D integer array C, it should be initialized to zero. You should assume that the first row of C starts from 1, and the first column of C starts from 'A'.

void Set(int row, char column, int val): Change the value at C(row, column) to be val.

int Get(int row, char column): Return the value at C(row, column).

int Sum(int row, char column, List of Strings : numbers): This function calculate and set the value at C(row, column), where the value should be the sum of cells represented by numbers. This function return the sum result at C(row, column). This sum formula should exist until this cell is overlapped by another value or another sum formula.

numbers is a list of strings that each string represent a cell or a range of cells. If the string represent a single cell, then it has the following format : ColRow. For example, "F7" represents the cell at (7, F).

If the string represent a range of cells, then it has the following format : ColRow1:ColRow2. The range will always be a rectangle, and ColRow1 represent the position of the top-left cell, and ColRow2 represents the position of the bottom-right cell.

Example 1:

Excel(3,"C");
// construct a 3*3 2D array with all zero.
//   A B C
// 1 0 0 0
// 2 0 0 0
// 3 0 0 0

Set(1, "A", 2);
// set C(1,"A") to be 2.
//   A B C
// 1 2 0 0
// 2 0 0 0
// 3 0 0 0

Sum(3, "C", ["A1", "A1:B2"]);
// set C(3,"C") to be the sum of value at C(1,"A") and the values sum of the rectangle range whose top-left cell is C(1,"A") and bottom-right cell is C(2,"B"). Return 4.
//   A B C
// 1 2 0 0
// 2 0 0 0
// 3 0 0 4

Set(2, "B", 2);
// set C(2,"B") to be 2. Note C(3, "C") should also be changed.
//   A B C
// 1 2 0 0
// 2 0 2 0
// 3 0 0 6

Note:

1. You could assume that there won't be any circular sum reference. For example, A1 = sum(B1) and B1 = sum(A1).
2. The test cases are using double-quotes to represent a character.
3. Please remember to RESET your class variables declared in class Excel, as static/class variables are persisted across multiple test cases. Please see here for more details.

这道题让我们设计Excel表格的求和公式，Excel表格想必大家都用过，还是比较熟悉的，这里让我们对单元格进行求和运算。由于这道题里要求二维数组的局部和，而且又会经常更新数组的值，博主第一反应觉得应该用之前那题Range Sum Query 2D - Mutable中的树状数组来做，结果哼哼哧哧的写完后，发现下面这个test case没通过：

["Excel","sum","set","get"]
[[3,"C"],[1,"A",["A2"]],[2,"A",1],[1,"A"]]
Expected:
[null,0,null,1]

仔细分析一下发现，这个case先把A2的值赋给了A1，此时A1和A2都是0，然后给A2赋值为1，求A1的值。大家的第一印象肯定是觉得A1还是0啊，其实在Excel中，相当于已经把A1和A2关联起来了，只要A2点值发生了改变，A1的值也会跟着变，所以A1的值此时也为1。而树状数组的主要功能的优化区域和的计算速度，并没有建立关联的步骤，难怪不能通过OJ呢。这道题标记为Hard还是有道理的，我们要模拟出Excel表中的这种关联方式，这里参考的是yupinglu大神的帖子，首先我们肯定需要一个二维数组mat来保存数据，然后需要一个map来建立单元格和区域和之间的映射，这里的区域和就是sum函数中的字符串数组表示的内容，可参见题目中的例子，有可能单个单元格或者多个。

我们来看set函数，如果我们改变了某个单元格的内容，那么如果作为结果单元格，那么对应的链接就会断开。比如我们有三个单元格A1, B1, C1，我们设置的关联是A1 + B1 = C1，那么我们改变A1和B1的值都是OK的，C1的值会自动更新。但如果我们改变了C1的值，那么这个关联就不复存在了，Excel中也是这样的。所以我们在改变某个单元格的时候，要将其的关联删除。

我们再来看get函数，我们在获取某个单元格的值的时候，一定要先看其有没有和其他单元格关联，如果有的话，要重新计算一下关联，有可能关联的单元格的值已经发生改变了，那么当前作为结果单元格的值也需要改变；如果该单元格没有任何关联，那么就直接从数组mat中取值即可。

最后看本题的难点sum函数，要根据关联格求出结果格的值，首先这个字符串数组可能有多个字符串，每个字符串有两个可能，一种是单个的单元格，一种是两个单元格中间用冒号隔开。那么我们需要分情况讨论，区别这两种情况的方法就是看冒号是否存在，如果不存在，就说明只有一个单元格，我们将其数字和字母都提取出来，调用get函数，将该位置的值加入结果res中；如果冒号存在，我们根据冒号的位置，分别将两个单元格的字母和数字提取出来，然后遍历这两个单元格之间所有的单元格，调用get函数并将返回值加入结果res中。这个遍历相加的过程可能可以用树状数组来优化，但由于这不是此题的考察重点，所以直接遍历就OK。最后别忘了建立目标单元格和区域字符串数组之间的映射，并返回结果res即可。

class Excel {
public:
    Excel(int H, char W) {
        m.clear();
        mat.resize(H, vector<int>(W - 'A', ));
    }

    void set(int r, char c, int v) {
        if (m.count({r, c})) m.erase({r, c});
        mat[r - ][c - 'A'] = v;
    }

    int get(int r, char c) {
        if (m.count({r, c})) return sum(r, c, m[{r, c}]);
        return mat[r - ][c - 'A'];
    }

    int sum(int r, char c, vector<string> strs) {
        int res = ;
        for (string str : strs) {
            auto found = str.find_last_of(":");
            if (found == string::npos) {
                char y = str[];
                int x = stoi(str.substr());
                res += get(x, y);
            } else {
                int x1 = stoi(str.substr(, (int)found - )), y1 = str[] - 'A';
                int x2 = stoi(str.substr(found + )), y2 = str[found + ] - 'A';
                for (int i = x1; i <= x2; ++i) {
                    for (int j = y1; j <= y2; ++j) {
                        res += get(i, j + 'A');
                    }
                }
            }
        }
        m[{r, c}] = strs;
        return res;
    }

private:
    vector<vector<int>> mat;
    map<pair<int, char>, vector<string>> m;
};

参考资料：

https://discuss.leetcode.com/topic/93819/c-3-ms-solution-easy-to-understand

https://discuss.leetcode.com/topic/93812/c-3-ms-concise-and-easy-to-understand

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