[Leetcode] 73. Set Matrix Zeroes 解题报告
2017-01-30 04:01
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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.
click to show follow up.
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?
思路:
这道题目的难点在于如何只使用常数量级的空间复杂度,其技巧是在已有数据上做文章:1)检查并记录第一行和第一列是否需要归零;2)遍历矩阵,一旦发现某个元素为0,则将它所在的行和列的第一个元素置为0;3)遍历第一行和第一列,如果是0,则将其对应的整个行或者整个列的元素置为0;4)如果第一行或者第一列需要归零,则将其归零。时间复杂度为O(m*n),空间复杂度为O(1)。
代码:
class Solution {
public:
void setZeroes(vector<vector<int>>& matrix) {
bool zeroFirstRow = false, zeroFirstCol = false; // step 1: mark first row and first col
for(int i = 0; i < matrix[0].size(); ++i)
if(matrix[0][i] == 0)
{
zeroFirstRow = true;
break;
}
for(int i = 0; i < matrix.size(); ++i)
if(matrix[i][0] == 0)
{
zeroFirstCol = true;
break;
}
for(int i = 1; i < matrix.size(); ++i) // step 2: mark all rows and cols that should be set zero
{
for(int j = 1; j < matrix[i].size(); ++j)
{
if(matrix[i][j] == 0)
{
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
for(int i = 1; i < matrix.size(); ++i) // step 3: set zero of other rows and cols
{
for(int j = 1; j < matrix[i].size(); ++j)
{
4000
if(matrix[i][0] == 0 || matrix[0][j] == 0)
matrix[i][j] = 0;
}
}
if(zeroFirstRow) // step 4: set zero of first row and col if necessary
{
for(int i = 0; i < matrix[0].size(); ++i)
matrix[0][i] = 0;
}
if(zeroFirstCol)
{
for(int i = 0; i < matrix.size(); ++i)
matrix[i][0] = 0;
}
}
};
Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
click to show follow up.
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?
思路:
这道题目的难点在于如何只使用常数量级的空间复杂度,其技巧是在已有数据上做文章:1)检查并记录第一行和第一列是否需要归零;2)遍历矩阵,一旦发现某个元素为0,则将它所在的行和列的第一个元素置为0;3)遍历第一行和第一列,如果是0,则将其对应的整个行或者整个列的元素置为0;4)如果第一行或者第一列需要归零,则将其归零。时间复杂度为O(m*n),空间复杂度为O(1)。
代码:
class Solution {
public:
void setZeroes(vector<vector<int>>& matrix) {
bool zeroFirstRow = false, zeroFirstCol = false; // step 1: mark first row and first col
for(int i = 0; i < matrix[0].size(); ++i)
if(matrix[0][i] == 0)
{
zeroFirstRow = true;
break;
}
for(int i = 0; i < matrix.size(); ++i)
if(matrix[i][0] == 0)
{
zeroFirstCol = true;
break;
}
for(int i = 1; i < matrix.size(); ++i) // step 2: mark all rows and cols that should be set zero
{
for(int j = 1; j < matrix[i].size(); ++j)
{
if(matrix[i][j] == 0)
{
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
for(int i = 1; i < matrix.size(); ++i) // step 3: set zero of other rows and cols
{
for(int j = 1; j < matrix[i].size(); ++j)
{
4000
if(matrix[i][0] == 0 || matrix[0][j] == 0)
matrix[i][j] = 0;
}
}
if(zeroFirstRow) // step 4: set zero of first row and col if necessary
{
for(int i = 0; i < matrix[0].size(); ++i)
matrix[0][i] = 0;
}
if(zeroFirstCol)
{
for(int i = 0; i < matrix.size(); ++i)
matrix[i][0] = 0;
}
}
};
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