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(m__n) 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?
代码语言：txt复制 public void setZeroes(int[][] matrix) {
if (matrix == null || matrix.length == 0) {
return;
}
int mx = matrix.length;
int my = matrix[0].length;
// 两个变量，判断第一行和第一列是否有0
boolean xflag = false, yflag = false;
// 判断第一行是否有0
for (int i = 0; i < mx; i ) {
if (matrix[i][0] == 0) {
xflag = true;
break;
}
}
// 判断第一列是否有0
for (int i = 0; i < my; i ) {
if (matrix[0][i] == 0) {
yflag = true;
break;
}
}
// 其它行、列是否有0
for (int i = 1; i < mx; i ) {
for (int j = 1; j < my; j ) {
if (matrix[i][j] == 0) {
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
// 对于第一列，为0，则将所在行变成0
for (int i = 1; i < mx; i ) {
if (matrix[i][0] == 0) {
for (int j = 0; j < my; j ) {
matrix[i][j] = 0;
}
}
}
// 对于第一行，为0，则将所在列变成0
for (int i = 0; i < my; i ) {
if (matrix[0][i] == 0) {
for (int j = 0; j < mx; j ) {
matrix[j][i] = 0;
}
}
}
// 若原来第一行中有0，则将整行置0
if (xflag) {
for (int i = 0; i < mx; i ) {
matrix[i][0] = 0;
}
}
// 若原来第一列中有0，则将整列置0
if (yflag) {
for (int i = 0; i < my; i ) {
matrix[0][i] = 0;
}
}
}
