| # |
Title |
Difficulty |
Topic |
| 73 |
Set Matrix Zeroes |
Medium |
Array |
Description
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:
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Input:
[
[1,1,1],
[1,0,1],
[1,1,1]
]
Output:
[
[1,0,1],
[0,0,0],
[1,0,1]
]
|
Example 2:
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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(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?
Analyze
题目要求$$O(1)$$的空间复杂度。如果$$i$$行$$j$$列的元素为0，则设$$i$$行第一个元素和$$j$$列第一个元素为0。但是，这样会将$$i$$行0列和0行$$j$$列的信息抹除，因此还需要特别储存$$i$$行0列和0行$$j$$列是否有0。
Submission
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class {
public void setZeroes(int[][] matrix) {
int m = matrix.length, n = matrix[0].length;
boolean iZero = false, jZero = false;
for(int i=0; i<m; i++) {
for(int j=0; j<n; j++) {
if(matrix[i][j] == 0) {
if(i == 0) iZero = true;
if(j == 0) jZero = true;
matrix[i][0] = 0;
matrix[0][j] = 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(iZero) {
for(int j=0; j<n; j++)
matrix[0][j] = 0;
}
if(jZero) {
for(int i=0; i<m; i++)
matrix[i][0] = 0;
}
}
}
|
Complex Analyze
时间复杂度为$$O(m times n)$$，空间复杂度$$O(1)$$。
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