JavaScript 如何根据矩阵元素的方向标记移动重排元素

初始定义的矩阵 #

let m = [
  ["0",      "0",      "S",       "S",       "0",   "0"],
  ["0",      "S", "D^left", "D^right",       "S",   "0"],
  ["S", "D^left", "Dvleft", "Dvright", "D^right",   "S"],
  ["S", "Dvleft",   "D<up",    "D>up", "Dvright",   "S"],
  ["0",      "S", "D<down",  "D>down",       "S",   "0"],
  ["0",      "0",      "S",       "S",       "0",   "0"]
];

重排规则 #

• 所有带方向标记的元素，按照标记指定的方向移动1格
• 原矩阵中值为"0"的位置保持固定，不被其他元素覆盖
• 其余未被固定0、移动后元素覆盖的位置，统一填充为"S"

期望输出矩阵 #

m = [
  ["0",      "0",      "S",      "S",       "0",       "0"],
  ["0", "D^left",      "S",      "S", "D^right",       "0"],
  ["D^left", "Dvleft", "D<up", "D>up", "Dvright", "D^right"],
  ["Dvleft",      "S",      "S",      "S",       "S", "Dvright"],
  ["0",      "S",      "S",      "S",       "S",       "0"],
  ["0",      "0", "D<down", "D>down",       "0",       "0"]
];

原实现存在的问题 #

直接在原矩阵上边遍历边修改元素位置的方式，会引发两个核心问题：

• 已经移动到新位置的元素，会在后续遍历到该位置时被重复识别，按照自身方向再次移动，导致最终位置偏移
• 原位置的填充操作会覆盖其他尚未遍历到的待移动元素，造成元素丢失、位置错乱

实现思路 #

1. 新建与原矩阵尺寸完全一致的结果矩阵，先遍历原矩阵，把所有固定的"0"位置原样复制到结果矩阵，其余位置初始填充为"S"
2. 遍历原矩阵的每个元素，对带方向标记的元素，根据标记内容计算目标位置的行列偏移量，直接将元素写入结果矩阵的对应位置
3. 整个过程不修改原矩阵内容，从根源上避免遍历过程中的状态干扰

参考实现代码 #

function rearrangeMatrix(matrix) {
  const rowCount = matrix.length;
  const colCount = matrix[0].length;
  // 初始化结果矩阵，默认填充"S"
  const result = Array.from({length: rowCount}, () => Array(colCount).fill("S"));
  // 复制固定的"0"位置
  for (let row = 0; row < rowCount; row++) {
    for (let col = 0; col < colCount; col++) {
      if (matrix[row][col] === "0") {
        result[row][col] = "0";
      }
    }
  }
  // 处理带方向标记的元素，写入目标位置
  for (let row = 0; row < rowCount; row++) {
    for (let col = 0; col < colCount; col++) {
      const item = matrix[row][col];
      if (!item.startsWith("D")) continue;
      let targetRow = row;
      let targetCol = col;
      // 根据标记计算偏移量
      if (item.includes("left")) targetCol -= 1;
      if (item.includes("right")) targetCol += 1;
      if (item.includes("up")) targetRow -= 1;
      if (item.includes("down")) targetRow += 1;
      // 边界校验，确保目标位置有效
      if (targetRow >= 0 && targetRow < rowCount && targetCol >=0 && targetCol < colCount) {
        result[targetRow][targetCol] = item;
      }
    }
  }
  return result;
}
// 执行重排
const newMatrix = rearrangeMatrix(m);

内容的提问来源于stack exchange，提问作者user434002