54 / 59 Spiral Matrix( I / II )

Total Accepted: 61182 Total Submissions: 268396 Difficulty: Medium

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].

分析：

控制不好邊界條件，自己寫的那叫一個亂！

鑑賞別人的算法，那叫一個漂亮！

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> result;
        if(matrix.empty())
            return result;
        //這是思路最顯然，最漂的方式，參考討論區
        int rowBegin = 0;//行開始
        int rowEnd = matrix.size()-1;//行結束
        int colBegin = 0;//列開始
        int colEnd = matrix[0].size() - 1;//列結束

        while (rowBegin <= rowEnd && colBegin <= colEnd) {
            // 從左到右
            for (int j = colBegin; j <= colEnd; j ++)
                result.push_back(matrix[rowBegin][j]);
            rowBegin++;//收縮行開始

            // 從上到下
            for (int j = rowBegin; j <= rowEnd; j ++) 
                result.push_back(matrix[j][colEnd]);
            colEnd--;
            
            // 從右到左
            if (rowBegin <= rowEnd) {
                for (int j = colEnd; j >= colBegin; j --)
                    result.push_back(matrix[rowEnd][j]);
            }
            rowEnd--;

            // 從下到上
            if (colBegin <= colEnd) {
                for (int j = rowEnd; j >= rowBegin; j --)
                    result.push_back(matrix[j][colBegin]);
            }
            colBegin++;
        }

        return result;
    }
};

59. Spiral Matrix II

Total Accepted: 54401 Total Submissions: 153690 Difficulty: 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 ]
]

分析：

和上一題一樣

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector<vector<int>> matrix;
        vector<int>  aline(n,0);
        for(int i=0;i<n;i++)
            matrix.push_back(aline);
        int rowBegin = 0;//行開始  
        int rowEnd = n-1;//行結束  
        int colBegin = 0;//列開始  
        int colEnd = n-1;//列結束  
        int val=1;
        while (rowBegin <= rowEnd && colBegin <= colEnd) {  
            // 從左到右  
            for (int j = colBegin; j <= colEnd; j ++)  
                matrix[rowBegin][j]= val++;
            rowBegin++;//收縮行開始  
  
            // 從上到下  
            for (int j = rowBegin; j <= rowEnd; j ++)   
                matrix[j][colEnd] = val++; 
            colEnd--;  
              
            // 從右到左  
            if (rowBegin <= rowEnd) {  
                for (int j = colEnd; j >= colBegin; j --)  
                    matrix[rowEnd][j]=val++;
            }  
            rowEnd--;  
  
            // 從下到上  
            if (colBegin <= colEnd) {  
                for (int j = rowEnd; j >= rowBegin; j --)  
                    matrix[j][colBegin]=val++;  
            }  
            colBegin++;  
        }  
  
        return matrix;      
    }
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/51585437
原作者博客：http://blog.csdn.net/ebowtang
本博客LeetCode題解索引：http://blog.csdn.net/ebowtang/article/details/50668895