Algorithm: Array(1)

54. Spiral Matrix

59. Spiral Matrix II

73. Set Matrix Zeroes

54. Spiral Matrix

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.


For example,
Given the following matrix:


[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]

You should return [1,2,3,6,9,8,7,4,5].

思路：这道题的pattern是利用int row; int col; 接着每次就用 for (int i = 0; i < row; i++)去填充某一个行、某一列之后，row--（刚刚填充了一行）或者col--（刚刚填充了一列）。接着判断停止的标准是row == 0或者 col == 0。注意是填充的变量，因为是优先填充第一行，所以col填充变量要从-1开始。如果优先填充一列，row填充变量从0开始。

public class Solution {    public List<Integer> spiralOrder(int[][] matrix) {        List<Integer> ans = new ArrayList<>();        if (matrix == null || matrix.length == 0            || matrix[0] == null || matrix[0].length == 0) {                return ans;        }                int row = matrix.length;        int col = matrix[0].length;                int rowPos = 0;        int colPos = -1;                while (true) {            for (int t = 0; t < col; t++) {                ans.add(matrix[rowPos][++colPos]);            }            if (--row == 0) {                break;            }            // 因为上面已经自减了，所以是row = row - 1了。            for (int t = 0; t < row; t++) {                ans.add(matrix[++rowPos][colPos]);            }            if (--col == 0) {                break;            }            for (int t = 0; t < col; t++) {                ans.add(matrix[rowPos][--colPos]);            }            if (--row == 0) {                break;            }            for (int t = 0; t < row; t++) {                ans.add(matrix[--rowPos][colPos]);            }            if (--col == 0) {                break;            }        }                return ans;    }}

59. Spiral Matrix II


Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.


For example,
Given n = 3,


You should return the following matrix:
[
 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
]

思路与上题一致。

public class Solution {    public int[][] generateMatrix(int n) {        if (n <= 0) {            return new int[0][0];        }                int[][] res = new int[n][n];                int row = 0;        int col = -1;                int colNum = n;        int rowNum = n;                int num = 0;                while (true) {            for (int t = 0; t < colNum; t++) {                res[row][++col] = ++num;            }            if (--rowNum == 0) {                break;            }            for (int t = 0; t < rowNum; t++) {                res[++row][col] = ++num;            }            if (--colNum == 0) {                break;            }            for (int t = 0; t < colNum; t++) {                res[row][--col] = ++num;            }            if (--rowNum == 0) {                break;            }            for (int t = 0; t < rowNum; t++) {                res[--row][col] = ++num;            }            if (--colNum == 0) {                break;            }        }                return res;    }}

73. Set Matrix Zeroes

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：如代码comment. 同时，需要考虑的是，在最后动到第0行或者第0列之前，任何操作都不能动到第0行或者第0列。

public class Solution {    // 题目的最简单方法如下：    // 利用第0行和第0列作为某行某列是否应该设为0的依据。因为，在同一行当中，任何列为0，正行为0。所以，第0列或者第n列为0的效果    // 是一样的。所以，利用这个原理，第0行和第0列单独处理记录。接着扫描剩余的所有格。一旦一个格子为0，就把对应的0行、0列的元    // 素置为零。    // 然后根据扫描后的零行零列记录，置对应的列和行为0后。根据原来0行、0列的记录，给对应的0行、0列置零。    public void setZeroes(int[][] matrix) {        if (matrix == null || matrix.length == 0            || matrix[0] == null || matrix[0].length == 0) {                return;            }                boolean rowFlag = false;        boolean colFlag = false;                int colNum = matrix[0].length;        int rowNum = matrix.length;                for (int i = 0; i < colNum; i++) {            if (matrix[0][i] == 0) {                rowFlag = true;                break;            }        }                for (int j = 0; j < rowNum; j++) {            if (matrix[j][0] == 0) {                colFlag = true;                break;            }        }                for (int i = 1; i < rowNum; i++) {            for (int j = 1; j < colNum; j++) {                if (matrix[i][j] == 0) {                    matrix[0][j] = 0;                    matrix[i][0] = 0;                }            }        }                for (int i = 1; i < colNum; i++) {//注意是从1开始，不能碰第0列            if (matrix[0][i] == 0) {                for (int k = 1; k < rowNum; k++) {                    matrix[k][i] = 0;                }            }        }                for (int j = 1; j < rowNum; j++) {//注意是从1开始，不能碰第0行            if (matrix[j][0] == 0) {                for (int k = 1; k < colNum; k++) {                    matrix[j][k] = 0;                }            }        }                //返回处理第0行，第0列        if (rowFlag) {            for (int k = 0; k < colNum; k++) {                matrix[0][k] = 0;            }        }                if (colFlag) {            for (int k = 0; k < rowNum; k++) {                matrix[k][0] = 0;            }        }    }}