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].

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  (M) Spiral Matrix II

分析：

控制不好边界条件，被此题虐了！鉴赏别人的算法，那叫一个漂亮！

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> result;
        if(matrix.empty())
            return result;
        //这是思路最显然，最漂的方式，参考讨论区
        int rowBegin = 0;//行开始
        int rowEnd = matrix.size()-1;//行结束
        int colBegin = 0;//列开始
        int colEnd = matrix[0].size() - 1;//列结束

        while (rowBegin <= rowEnd && colBegin <= colEnd) {
            // 从左到右
            for (int j = colBegin; j <= colEnd; j ++)
                result.push_back(matrix[rowBegin][j]);
            rowBegin++;//收缩行开始

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

            // 从下到上
            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 ]
]

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分析：

和上一题一样

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;//行开始  
        int rowEnd = n-1;//行结束  
        int colBegin = 0;//列开始  
        int colEnd = n-1;//列结束  
        int val=1;
        while (rowBegin <= rowEnd && colBegin <= colEnd) {  
            // 从左到右  
            for (int j = colBegin; j <= colEnd; j ++)  
                matrix[rowBegin][j]= val++;
            rowBegin++;//收缩行开始  
  
            // 从上到下  
            for (int j = rowBegin; j <= rowEnd; j ++)   
                matrix[j][colEnd] = val++; 
            colEnd--;  
              
            // 从右到左  
            if (rowBegin <= rowEnd) {  
                for (int j = colEnd; j >= colBegin; j --)  
                    matrix[rowEnd][j]=val++;
            }  
            rowEnd--;  
  
            // 从下到上  
            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题解索引：http://blog.csdn.net/ebowtang/article/details/50668895