标签:bad pen first any tab question abs problems follow
https://leetcode.com/problems/set-matrix-zeroes/tabs/description
Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
Did you use extra space?
A straight forward solution using O(mn) space is probably a bad idea.
A simple improvement uses O(m + n) space, but still not the best solution.
Could you devise a constant space solution?
class Solution(object): def setZeroes(self, matrix): """ :type matrix: List[List[int]] :rtype: void Do not return anything, modify matrix in-place instead. """ # Space O(m+n) # remember index of rows and columns with zero. rows = [] cols = [] rowNum = len(matrix) colNum = len(matrix[0]) # detect zeors and remember their row and col index. for i in range(rowNum): for j in range(colNum): if matrix[i][j] == 0: rows.append(i) cols.append(j) # set marked cols and rows to zeros for row in rows: for col in range(colNum): matrix[row][col] = 0 for col in cols: for row in range(rowNum): matrix[row][col] = 0
Sol 2:
Java
public class Solution { public void setZeroes(int[][] matrix) { // Time O(m*n) Space O(1) final int m = matrix.length; final int n = matrix[0].length; boolean row_has_zero = false; // check if the first row has 0 boolean col_has_zero = false; // check if the first col has 0 for (int i = 0; i < n; i++) if (matrix[0][i] == 0) { row_has_zero = true; break; } for (int i = 0; i < m; i++) if (matrix[i][0] == 0) { col_has_zero = true; break; } for (int i = 1; i < m; i++) for (int j = 1; j < n; j++) if (matrix[i][j] == 0) { matrix[0][j] = 0; matrix[i][0] = 0; } for (int i = 1; i < m; i++) for (int j = 1; j < n; j++) if (matrix[i][0] == 0 || matrix[0][j] == 0) matrix[i][j] = 0; if (row_has_zero) for (int i = 0; i < n; i++) matrix[0][i] = 0; if (col_has_zero) for (int i = 0; i < m; i++) matrix[i][0] = 0; } };
标签:bad pen first any tab question abs problems follow
原文地址:http://www.cnblogs.com/prmlab/p/7246042.html