<LeetCode OJ> 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