Algorithm-Matrix Traversal Problem Album
1. Spiral matrix
给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。
示例 1:
输入:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
输出: [1,2,3,6,9,8,7,4,5]
示例 2:
输入:
[
[1, 2, 3, 4],
[5, 6, 7, 8],
[9,10,11,12]
]
输出: [1,2,3,4,8,12,11,10,9,5,6,7]
This problem is actually very simple. The key to the solution is how to determine the boundary conditions of the traversal.
Regarding the matrix as an onion, we poke the heart of the matrix layer by layer. That is, the layers of the matrix.
Define two variables: the current layer, the number of remaining unprinted elements count as boundary conditions
We then traverse the four directions from left to right, east to bottom, right to left, and bottom to top to complete the layer number condition. Each time we print, we subtract one element. When the element is reduced to 0, it ends the traversal
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> list=new ArrayList<>();
if(matrix.length==0){
return list;
}
int m=matrix.length-1;
int n=matrix[0].length-1;
int count=(m+1)*(n+1);
int layer=0;
while (count>0){
//从左到右
for (int i=layer;i<=n-layer&&count>0;i++){
list.add(matrix[layer][i]);
count--;
}
//从上到下
for (int i=layer+1;i<=m-layer&&count>0;i++){
list.add(matrix[i][n-layer]);
count--;
}
//从右到左
for (int i=n-layer-1;i>=layer&&count>0;i--){
list.add(matrix[m-layer][i]);
count--;
}
//从下到上
for (int i=m-layer-1;i>layer&&count>0;i--){
list.add(matrix[i][layer]);
count--;
}
layer++;//读完一层，继续下一层
}
return list;
}
2. Rotate the image
First look at the title description:
给定一个 n × n 的二维矩阵表示一个图像。
将图像顺时针旋转 90 度。
说明：
你必须在原地旋转图像，这意味着你需要直接修改输入的二维矩阵。请不要使用另一个矩阵来旋转图像。
示例 1:
给定 matrix =
[
[1,2,3],
[4,5,6],
[7,8,9]
],
原地旋转输入矩阵，使其变为:
[
[7,4,1],
[8,5,2],
[9,6,3]
]
示例 2:
给定 matrix =
[
[ 5, 1, 9,11],
[ 2, 4, 8,10],
[13, 3, 6, 7],
[15,14,12,16]
],
原地旋转输入矩阵，使其变为:
[
[15,13, 2, 5],
[14, 3, 4, 1],
[12, 6, 8, 9],
[16, 7,10,11]
]
I feel that this question is more like a math question. There are two solutions to this question.
The first type: read the elements clockwise into an array, and then adjust the array elements to fill in the
second type: take the transpose of the matrix, and then invert by row
The first way of writing is more cumbersome, here is the second way of writing
public void rotate(int[][] matrix) {
int n=matrix.length;
//矩阵转置
for(int i=0;i<n;i++){
for(int j=0;j<=i;j++){
int temp=matrix[i][j];
matrix[i][j]=matrix[j][i];
matrix[j][i]=temp;
}
}
//按行反转
for(int i=0;i<n;i++){
int left=0,right=n-1;
while(left<right){
int temp=matrix[i][left];
matrix[i][left]=matrix[i][right];
matrix[i][right]=temp;
left++;
right--;
}
}
}
3. Search for a two-dimensional matrix II
编写一个高效的算法来搜索 m x n 矩阵 matrix 中的一个目标值 target。该矩阵具有以下特性：
每行的元素从左到右升序排列。
每列的元素从上到下升序排列。
示例:
现有矩阵 matrix 如下：
[
[1, 4, 7, 11, 15],
[2, 5, 8, 12, 19],
[3, 6, 9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]
给定 target = 5，返回 true。
给定 target = 20，返回 false。
The solution to this problem is a bit clever, essentially a pruning method. Start searching from the upper right corner, compare the value in the upper right corner with the target value, crop one row or column each time, the complexity is O(m+n)
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix.length==0){
return false;
}
int right=matrix[0].length-1,top=0;//右上角
while(right>=0&&top<matrix.length){
if(matrix[top][right]>target){
right--;//裁掉一列
}else if(matrix[top][right]<target){
top++;//裁掉一行
}else{
return true;
}
}
return false;
}
4. Search for a two-dimensional matrix
编写一个高效的算法来判断 m x n 矩阵中，是否存在一个目标值。该矩阵具有如下特性：
每行中的整数从左到右按升序排列。
每行的第一个整数大于前一行的最后一个整数。
示例 1:
输入:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 3
输出: true
示例 2:
输入:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 13
输出: false
These two questions are the same, just apply it directly
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix.length==0){
return false;
}
int right=matrix[0].length-1,top=0;//右上角
while(right>=0&&top<matrix.length){
if(matrix[top][right]>target){
right--;//裁掉一列
}else if(matrix[top][right]<target){
top++;//裁掉一行
}else{
return true;
}
}
return false;
}
Efficiency is not bad
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