[LeetCode]Spiral Matrix

题目：Spiral Matrix

螺旋输出一个数组。

思路：

螺旋的方式遍历这个数组，一次外循环，遍历数组一圈，直到全部遍历完。

/***************************************************************************************************
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].
***************************************************************************************************/
#include<stdio.h>

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* spiralOrder(int** matrix, int matrixRowSize, int matrixColSize) {
    int *array = (int *)malloc(matrixRowSize*matrixColSize*sizeof(int));
    int rowUp = 0,rowLow = matrixColSize - 1,columnUp = 0,columnLow = matrixRowSize - 1;
    int i,index = 0;
    while(rowUp < rowLow && columnUp < columnLow){
        for(i = rowUp;i < rowLow;i++){//行的上下界
            array[index++] = matrix[columnUp][i];
        }
        for(i = columnUp;i < columnLow;i++){//列的上下界
            array[index++] = matrix[i][rowLow];
        }
        for(i = rowLow;i > rowUp;i--){
            array[index++] = matrix[columnLow][i];
        }
        for(i = columnLow;i > columnUp;i--){
            array[index++] = matrix[i][rowUp];
        }
        columnUp++;
        rowUp++;
        columnLow--;
        rowLow--;
      }
      if(rowUp == rowLow){//还剩一列
          for(i = columnUp;i <= columnLow;i++)
              array[index++] = matrix[i][rowUp];
      }else if(columnUp == columnLow){//还剩一行
          for(i = rowUp;i <= rowLow;i++)
              array[index++] = matrix[columnUp][i];
      }
      return array;
}

void main(){
    int num = 5;
    int **a = (int **)malloc(num*sizeof(int *));
    for(int i = 0;i < num;i++){
        a[i] = (int *)malloc(num*sizeof(int));
        for(int j = 0;j < num;j++){
            a[i][j] = i*num + j + 1;
            printf("%d ",a[i][j]);
        }
        printf("
");
    }
    int *ret = spiralOrder(a,num,num);
    for(int i = 0;i < num*num;i++){
        printf("%d ",ret[i]);
    }
    free(ret);
    for(int i = 0;i < num;i++){
        free(a[i]);
    }
    free(a);
}

 题目：Spiral MatrixII

给定一个数字n，则有n^2个数字，螺旋的形式将这些数字分布到n单元的数组中。

思路：

和上面类似一圈为一个周期，但是这个数组是nxn的不会有行列不等的情况。

package com.example.medium;

/**
 * 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 ]
 *]
 * @author FuPing
 *
 */
public class GenerateMatrix {
    public static int[][] generateMatrix(int n){
        if(n < 0)return null;
        int [][]matrix = new int[n][n];
        int up = 0,low = n - 1,i,index = 1;//index数字的值
        while(up < low){
            for(i = up;i < low;i++){//up表示行标下界,low表示列标的上界
                matrix[up][i] = index++;
            }
            for(i = up;i < low;i++){
                matrix[i][low] = index++;
            }
            for(i = low;i > up;i--){
                matrix[low][i] = index++;
            }
            for(i = low;i > up;i--){
                matrix[i][up] = index++;
            }
            up++;
            low--;
        }
        if(up == low)matrix[up][up] = index++;//n为奇数时
        return matrix;
    }
    
    public static void main(String args[]){
        int n = 0;
        int [][]a = generateMatrix(n);
        for(int i = 0;i < n;i++){
            for(int j = 0;j < n;j++){
                System.out.print(a[i][j] + "  ");
            }
            System.out.println();
        }
    }
}