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 ]
]

 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
以这种方式进行循环即可。

public class Solution {
    public int[][] generateMatrix(int n) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int[][] result = new int[n][n];
        int a = 1;
        int top = 0;
        int bot = n - 1;
        int left = 0;
        int right = n - 1;
        while(top < bot && left < right){
            for(int i = left; i < right; i ++)
            {
                result[top][i] = a ++;
            }
            for(int i = top; i < bot; i ++)
            {
                result[i][right] = a ++;
            }
            for(int i = right; i > left; i --)
            {
                result[bot][i] = a ++;
            }
            for(int i = bot; i > top; i --)
            {
                result[i][left] = a ++;    
            }
            top ++;
            bot --;
            left ++;
            right --;
        }
        if(top == bot && left == right){//对于n 为奇数的情况。
            result[top][left] = a;
        }
        return result;
    }
}

 第三遍：

public class Solution {
    public int[][] generateMatrix(int n) {
        int[][] result = new int[n][n];
        if(n == 0) return result;
        int top = 0, bot = n - 1, num = 1;
        while(top < bot){
            for(int i = top; i < bot; i ++)
                result[top][i] = num ++;
            for(int i = top; i < bot; i ++)
                result[i][bot] = num ++;
            for(int i = bot; i > top; i --)
                result[bot][i] = num ++;
            for(int i = bot; i > top; i --)
                result[i][top] = num ++;
            top ++; bot --;
        }
        if(top == bot) result[top][bot] = num;
        return result;
    }
}