LeetCode: Spiral Matrix II 解题报告-三种方法解决旋转矩阵问题

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 ]
]

SOLUTION 1:

还是与上一题Spiral Matrix类似的算法，我们使用x1,y1作为左上角的起点，x2,y2记录右下角，这样子旋转时会简单多了。

 public int[][] generateMatrix1(int n) {
         int[][] ret = new int[n][n];

         if (n == ) {
             // return a [] not a NULL.
             return ret;
         }

         int number = ;
         int rows = n;

         int x1 = ;
         int y1 = ;

         while (rows > ) {
             int x2 = x1 + rows - ;
             int y2 = y1 + rows - ;

             // the Whole first row.
             for (int i = y1; i <= y2; i++) {
                 number++;
                 ret[x1][i] = number;
             }

             // the right column except the first and last line.
             for (int i = x1 + ; i < x2; i++) {
                 number++;
                 ret[i][y2] = number;
             }

             // This line is very important.
             if (rows <= ) {
                 break;
             }

             // the WHOLE last row.
             for (int i = y2; i >= y1; i--) {
                 number++;
                 ret[x2][i] = number;
             }

             // the left column. column keep stable
             // x: x2-1 --> x1 + 1
             for (int i = x2 - ; i > x1; i--) {
                 number++;
                 ret[i][y1] = number;
             }

             // remember this.
             rows -= ;
             x1++;
             y1++;
         }

         return ret;
     }

SOLUTION 2:

还是与上一题Spiral Matrix类似的算法，使用Direction 数组来定义旋转方向。其实蛮复杂的，也不好记。但是记住了应该是标准的算法。

 /*
         Solution 2: use direction.
     */
     public int[][] generateMatrix2(int n) {
         int[][] ret = new int[n][n];
         if (n == ) {
             return ret;
         }

         int[] x = {, , -, };
         int[] y = {, , , -};

         int num = ;

         int step = ;
         int candElements = ;

         int visitedRows = ;
         int visitedCols = ;

         // 0: right, 1: down, 2: left, 3: up.
         int direct = ;

         int startx = ;
         int starty = ;

         while (true) {
             if (x[direct] == ) {
                 // visit the Y axis
                 candElements = n - visitedRows;
             } else {
                 // visit the X axis
                 candElements = n - visitedCols;
             }

             if (candElements <= ) {
                 break;
             }

             // set the cell.
             ret[startx][starty] = ++num;
             step++;

             // change the direction.
             if (step == candElements) {
                 step = ;
                 visitedRows += x[direct] ==  ? : ;
                 visitedCols += y[direct] ==  ? : ;

                 // change the direction.
                 direct = (direct + ) % ;
             }

             startx += y[direct];
             starty += x[direct];
         }

         return ret;
     }

SOLUTION 3:

无比巧妙的办法，某人的男朋友可真是牛逼啊！[leetcode] Spiral Matrix | 把一个2D matrix用螺旋方式打印

此方法的巧妙之处是使用TOP,BOOTOM, LEFT, RIGHT 四个边界条件来限制访问。其实和第一个算法类似，但是更加简洁易懂。10分钟内AC!

 /*
         Solution 3: 使用四条bound来限制的方法.
     */
     public int[][] generateMatrix(int n) {
         int[][] ret = new int[n][n];
         if (n == ) {
             return ret;
         }

         int top = , bottom = n - , left = , right = n - ;
         int num = ;
         while (top <= bottom) {
             if (top == bottom) {
                 ret[top][top] = num++;
                 break;
             }

             // first line.
             for (int i = left; i < right; i++) {
                 ret[top][i] = num++;
             }

             // right line;
             for (int i = top; i < bottom; i++) {
                 ret[i][right] = num++;
             }

             // bottom line;
             for (int i = right; i > left; i--) {
                 ret[bottom][i] = num++;
             }

             // left line;
             for (int i = bottom; i > top; i--) {
                 ret[i][left] = num++;
             }

             top++;
             bottom--;
             left++;
             right--;
         }

         return ret;
     }

GitHub Code:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/array/GenerateMatrix1.java