566. Reshape the Matrix
2018-02-06 13:49
316 查看
In MATLAB, there is a very useful function called 'reshape', which can reshape a matrix into a new one with different size but keep its original data.
You're given a matrix represented by a two-dimensional array, and two
positive integers r and c representing the
row number and column number of the wanted reshaped matrix, respectively.
The reshaped matrix need to be filled with all the elements of the original matrix in the same
row-traversing order as they were.
If the 'reshape' operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.
Example 1:
Example 2:
Note:
The height and width of the given matrix is in range [1, 100].
The given r and c are all positive.
思路:对号入座
代码1:class Solution {
public:
vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
int o_r=nums.size(), o_c=nums[0].size();
if (!(o_r*o_c-r*c)){
vector<vector<int>> Matrix(r, vector<int>(c));// vector initial
for(int i=0;i<(r*c);i++){
Matrix[i/c][i%c]=nums[i/o_c][i%o_c];
}
return Matrix;
}
else{
return nums;
}
}
};
代码2:class Solution {
public:
vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
int o_r=nums.size(),oc=nums[0].size();
if(o_r*oc!=r*c){return nums;}
vector<int> inner(c,0);
vector<vector<int>> res(r,inner);
int ii=0,jj=0;
for(int i=0;i<o_r;i++){
for(int j=0;j<oc;j++){
res[ii][jj]=nums[i][j];
if(jj==c-1){
jj=0;
ii++;
}
else{jj++;}
}
}
return res;
}
};
You're given a matrix represented by a two-dimensional array, and two
positive integers r and c representing the
row number and column number of the wanted reshaped matrix, respectively.
The reshaped matrix need to be filled with all the elements of the original matrix in the same
row-traversing order as they were.
If the 'reshape' operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.
Example 1:
Input: nums = [[1,2], [3,4]] r = 1, c = 4 Output: [[1,2,3,4]] Explanation: The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.
Example 2:
Input: nums = [[1,2], [3,4]] r = 2, c = 4 Output: [[1,2], [3,4]] Explanation: There is no way to reshape a 2 * 2 matrix to a 2 * 4 matrix. So output the original matrix.
Note:
The height and width of the given matrix is in range [1, 100].
The given r and c are all positive.
思路:对号入座
代码1:class Solution {
public:
vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
int o_r=nums.size(), o_c=nums[0].size();
if (!(o_r*o_c-r*c)){
vector<vector<int>> Matrix(r, vector<int>(c));// vector initial
for(int i=0;i<(r*c);i++){
Matrix[i/c][i%c]=nums[i/o_c][i%o_c];
}
return Matrix;
}
else{
return nums;
}
}
};
代码2:class Solution {
public:
vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
int o_r=nums.size(),oc=nums[0].size();
if(o_r*oc!=r*c){return nums;}
vector<int> inner(c,0);
vector<vector<int>> res(r,inner);
int ii=0,jj=0;
for(int i=0;i<o_r;i++){
for(int j=0;j<oc;j++){
res[ii][jj]=nums[i][j];
if(jj==c-1){
jj=0;
ii++;
}
else{jj++;}
}
}
return res;
}
};
相关文章推荐
- 566. Reshape the Matrix重塑矩阵
- 566. Reshape the Matrix
- LeetCode 566. Reshape the Matrix (Easy)
- 566. Reshape the Matrix
- <LeetCode>566. Reshape the Matrix
- 566. Reshape the Matrix
- 566. Reshape the Matrix(Java)
- 566. Reshape the Matrix(2017/11/3)
- 566. Reshape the Matrix
- 566. Reshape the Matrix
- 566. Reshape the Matrix
- 566. Reshape the Matrix
- Leetcode 566. Reshape the Matrix 矩阵变形(数组,模拟,矩阵操作)
- Leetcode 566. Reshape the Matrix 矩阵变形(数组,模拟,矩阵操作)
- Leetcode 566. Reshape the Matrix(Easy)
- 566. Reshape the Matrix
- 566. Reshape the Matrix
- 566.Reshape the Matrix
- LeetCode-566-Reshape the Matrix-E
- leetcode 566: Reshape the Matrix