Set Matrix Zeroes leetcode java
题目:
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?
题解:
这道题是CC150 1.7上面的原题。可以看上面详尽的解释,我就不写了。
代码如下:
1 public void setZeroes(int[][] matrix) {
2 int m = matrix.length;
3 int n = matrix[0].length;
4
5 if(m==0||n==0)
6 return;
7 int[] flagr = new int[m];
8 int[] flagc = new int[n];
9
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
if(matrix[i][j]==0){
flagr[i]= 1;
flagc[j]= 1;
}
}
}
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
if(flagr[i]==1||flagc[j]==1){
matrix[i][j]=0;
}
}
}
}
另一种方法是不需要额外空间的,代码来自discuss:
1 public void setZeroes(int[][] matrix) {
2 int rownum = matrix.length;
3 if (rownum == 0) return;
4 int colnum = matrix[0].length;
5 if (colnum == 0) return;
6
7 boolean hasZeroFirstRow = false, hasZeroFirstColumn = false;
8
9 // Does first row have zero?
for (int j = 0; j < colnum; ++j) {
if (matrix[0][j] == 0) {
hasZeroFirstRow = true;
break;
}
}
// Does first column have zero?
for (int i = 0; i < rownum; ++i) {
if (matrix[i][0] == 0) {
hasZeroFirstColumn = true;
break;
}
}
// find zeroes and store the info in first row and column
for (int i = 1; i < matrix.length; ++i) {
for (int j = 1; j < matrix[0].length; ++j) {
if (matrix[i][j] == 0) {
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
// set zeroes except the first row and column
for (int i = 1; i < matrix.length; ++i) {
for (int j = 1; j < matrix[0].length; ++j) {
if (matrix[i][0] == 0 || matrix[0][j] == 0) matrix[i][j] = 0;
}
}
// set zeroes for first row and column if needed
if (hasZeroFirstRow) {
for (int j = 0; j < colnum; ++j) {
matrix[0][j] = 0;
}
}
if (hasZeroFirstColumn) {
for (int i = 0; i < rownum; ++i) {
matrix[i][0] = 0;
}
}
}
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