将嵌套的 while 循环更改为 for 循环... Javascript 矩阵

Question

大家好，任何人都可以将此 while 循环转换为 for 循环。我将非常感激，无论执行时间有多少，我都需要优化可读性，但我们希望避免复杂性并尽可能提高可读性，无需多言，这里是 sn-p。

let Matrix = [
  ["X", "X", null, null],
  [null, "X", "X", "X"],
  [null, null, "X", "X"],
  ["X", "X", null, null],
];
let rows = 4, cols = 4;
// this is making an array of 4 arrays inside of it filled with null as initial values
let OutputMatrix = [...new Array(rows).keys()].map((x) =>
  [...new Array(rows).keys()].map((i) => (i = i = null))
);
//i goes from last row to first row.
let counter = 1,
  i = rows - 1,
  j = 0,
  k = 1;
while (i >= 0) {
  while (i < rows && j < rows) {
    if (Matrix[i][j] != "X") {
      counter = 0;
    }
    OutputMatrix[i][j] = counter++;
    i++, j++;
  }
  j = 0;
  ++k;
  i = rows - k;
}

//j goes from second column to last column
(i = 0), (j = 1), (k = 1), (counter = 1);
while (j < cols) {
  while (i < rows && j < cols) {
    if (Matrix[i][j] != "X") {
      counter = 0;
    }
    OutputMatrix[i][j] = counter++;
    i++, j++;
  }
  i = 0;
  k++;
  j = k;
}

console.log(OutputMatrix);

输出

OutputMatrix = [
  [1, 1, 0, 0],
  [0, 2, 2, 1],
  [0, 0, 3, 3],
  [1, 1, 0, 0],
];

简而言之，这是从左上角到右下角的对角线，因此无论在哪里找到 "X"，它都会增加一，否则会发现 null flat counter 到零第一个嵌套while循环用于上半对角线，第二个用于其余部分

Answer 1

我认为您正在寻找这样的东西：

for (i = 0; i < rows; i++) {
    for (j = 0; j < cols; j++) {
        if (Matrix[i][j] == "X") {
            // If positioned on top row OR left column of matrix -> first X on diagonal
            if (i == 0 || j == 0) OutputMatrix[i][j] = 1;
            // Else: previous element on diagonal + 1
            else OutputMatrix[i][j] = OutputMatrix[i-1][j-1] + 1;
        }
        else OutputMatrix[i][j] = 0;
    }
}

这个 sn-p 将替换所有嵌套的 while 循环。