leetcode -- 73. Set Matrix Zeroes

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题目描述

题目难度：Medium
Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.

Example 1:

Input:
[
[1,1,1],
[1,0,1],
[1,1,1]
]
Output:
[
[1,0,1],
[0,0,0],
[1,0,1]
]

Example 2:

Input:
[
[0,1,2,0],
[3,4,5,2],
[1,3,1,5]
]
Output:
[
[0,0,0,0],
[0,4,5,0],
[0,3,1,0]
]
Follow up:

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?

AC代码1

Complexity Analysis

Time Complexity: O(M \times N)O(M×N) where M and N are the number of rows and columns respectively.

Space Complexity: O(M + N)O(M+N).

class Solution {public void setZeroes(int[][] matrix) {Set<Integer> rowSet = new HashSet<>();Set<Integer> colSet = new HashSet<>();int m = matrix.length;int n = matrix[0].length;for(int i = 0;i < m;i++){for(int j = 0;j < n;j++){if(matrix[i][j] == 0){rowSet.add(i);colSet.add(j);}}}setRowZeros(matrix, rowSet, n);setColZeros(matrix, colSet, m);}private void setRowZeros(int[][] matrix, Set<Integer> set, int n){for(Integer i : set)for(int j = 0;j < n;j++) matrix[i][j] = 0;}private void setColZeros(int[][] matrix, Set<Integer> set, int m){for(Integer i : set)for(int j = 0;j < m;j++) matrix[j][i] = 0;}
}

AC代码2

Complexity Analysis

Time Complexity : O(M + N)(M+N).
Space Complexity : O(1)

class Solution {public void setZeroes(int[][] matrix) {int MODIFIED = -1000000;int R = matrix.length;int C = matrix[0].length;for (int r = 0; r < R; r++) {for (int c = 0; c < C; c++) {if (matrix[r][c] == 0) {// We modify the corresponding rows and column elements in place.// Note, we only change the non zeroes to MODIFIEDfor (int k = 0; k < C; k++) {if (matrix[r][k] != 0) {matrix[r][k] = MODIFIED;}}for (int k = 0; k < R; k++) {if (matrix[k][c] != 0) {matrix[k][c] = MODIFIED;}}}}}for (int r = 0; r < R; r++) {for (int c = 0; c < C; c++) {// Make a second pass and change all MODIFIED elements to 0 """if (matrix[r][c] == MODIFIED) {matrix[r][c] = 0;}}}}
}

AC代码3

Complexity Analysis

Time Complexity : O(M×N)
Space Complexity : O(1)

class Solution {public void setZeroes(int[][] matrix) {Boolean isCol = false;int R = matrix.length;int C = matrix[0].length;for (int i = 0; i < R; i++) {// Since first cell for both first row and first column is the same i.e. matrix[0][0]// We can use an additional variable for either the first row/column.// For this solution we are using an additional variable for the first column// and using matrix[0][0] for the first row.if (matrix[i][0] == 0) {isCol = true;}for (int j = 1; j < C; j++) {// If an element is zero, we set the first element of the corresponding row and column to 0if (matrix[i][j] == 0) {matrix[0][j] = 0;matrix[i][0] = 0;}}}// Iterate over the array once again and using the first row and first column, update the elements.for (int i = 1; i < R; i++) {for (int j = 1; j < C; j++) {if (matrix[i][0] == 0 || matrix[0][j] == 0) {matrix[i][j] = 0;}}}// See if the first row needs to be set to zero as wellif (matrix[0][0] == 0) {for (int j = 0; j < C; j++) {matrix[0][j] = 0;}}// See if the first column needs to be set to zero as wellif (isCol) {for (int i = 0; i < R; i++) {matrix[i][0] = 0;}}}
}