题目链接:set-matrix-zeroes
import java.util.Arrays;
/**
*
Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
Follow up:
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?
*
*/
public class SetMatrixZeroes {
//解法一
// 157 / 157 test cases passed.
// Status: Accepted
// Runtime: 273 ms
// Submitted: 0 minutes ago
//时间复杂度O(mn) 空间复杂度O(1)
//复用第0行和第0列,用来标记某列和某行是否该置0
public void setZeroes(int[][] matrix) {
int m = matrix.length;
int n =matrix[0].length;
boolean row0 = false; //先标记第0行是否包含0
boolean col0 = false; //先标记第0列是否包含0
for(int i = 0; i < m; i++) {
if(matrix[i][0] == 0) {
col0 = true;
break;
}
}
for(int i = 0; i < n; i++) {
if(matrix[0][i] == 0) {
row0 = 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[0][j] == 0 || matrix[i][0] == 0 ) {
matrix[i][j] = 0;
}
}
}
if(col0) {
for(int i = 0; i < m; i++) {
matrix[i][0] = 0;
}
}
if(row0) {
for(int i = 0; i < n; i++) {
matrix[0][i] = 0;
}
}
}
//第二种解法
// 157 / 157 test cases passed.
// Status: Accepted
// Runtime: 365 ms
// Submitted: 0 minutes ago
//时间复杂度O(mn) 空间复杂度O(m+n)
public void setZeroes1(int[][] matrix) {
int m = matrix.length;
int n =matrix[0].length;
boolean[] row = new boolean[m];
boolean[] col = new boolean[n];
Arrays.fill(row, false);
Arrays.fill(col, false);
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if(matrix[i][j] == 0) {
row[i] = true;
col[j] = true;
}
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if(row[i] || col[j] ) {
matrix[i][j] = 0;
}
}
}
}
public static void main(String[] args) {
// TODO Auto-generated method stub
}
}[LeetCode 73] Set Matrix Zeroes
原文:http://blog.csdn.net/ever223/article/details/44603075