解题思路：检查并标记第一行和第一列是否有0，并使用变量标记使用第一行和第一列标记其余行列是否有0</l

复杂度：O(m*n) <div class="highlight highlight-source-js notranslate position-relative

✅73.矩阵置零 about like-algorithms HOT 2 OPEN

bazinga-web commented on June 21, 2024

✅73.矩阵置零

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Comments (2)

bazinga-web commented on June 21, 2024

解题思路：

检查并标记第一行和第一列是否有0，并使用变量标记
使用第一行和第一列标记其余行列是否有0
利用第一行第一列标0情况，将matrix中的数字标0
最后处理第一行和第一列

/**
 * @param {number[][]} matrix
 * @return {void} Do not return anything, modify matrix in-place instead.
 */
var setZeroes = function (matrix) {
  let firstRowHasZero = false;
  let firstColHasZero = false;

  for (let i = 0; i < matrix.length; i++) {
    if (matrix[i][0] === 0) {
      firstColHasZero = true;
    }
  }

  for (let i = 0; i < matrix[0].length; i++) {
    if (matrix[0][i] === 0) {
      firstRowHasZero = true;
    }
  }

  for (let i = 1; i < matrix.length; i++) {
    for (let j = 0; j < matrix[0].length; j++) {
      if (matrix[i][j] === 0) {
        matrix[i][0] = 0;
        matrix[0][j] = 0;
      }
    }
  }

  for (let i = 1; i < matrix.length; i++) {
    for (let j = 1; j < matrix[0].length; j++) {
      if (matrix[i][0] === 0 || matrix[0][j] === 0) {
        matrix[i][j] = 0;
      }
    }
  }

  if (firstColHasZero) {
    for (let i = 0; i < matrix.length; i++) {
      matrix[i][0] = 0;
    }
  }

  if (firstRowHasZero) {
    for (let i = 0; i < matrix[0].length; i++) {
      matrix[0][i] = 0;
    }
  }

  return matrix;
};

from like-algorithms.

Ray-56 commented on June 21, 2024

复杂度：O(m*n)

/**
 * @param {number[][]} matrix
 * @return {void} Do not return anything, modify matrix in-place instead.
 */
var setZeroes = function(matrix) {
	// O(n*m)
	const row = new Set();
	const column = new Set();

	for (let i = 0; i < matrix.length; i++) {
		for (let j = 0; j < matrix[0].length; j++) {
			if (matrix[i][j] === 0) {
				row.add(i);
				column.add(j);
			}
		}
	}

	for (let i = 0; i < matrix.length; i++) {
		for (let j = 0; j < matrix[0].length; j++) {
			if (row.has(i) || column.has(j)) {
				matrix[i][j] = 0;
			}
		}
	}
};