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LeetCode 59. 螺旋矩阵 II && LeetCode 54. 螺旋矩阵

Michael阿明

发布于 2021-02-20 06:12:59

32100

代码可运行

文章被收录于专栏：Michael阿明学习之路Michael阿明学习之路

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代码可运行

1. 题目信息

给定一个正整数 n，生成一个包含 1 到 n2 所有元素，且元素按顺时针顺序螺旋排列的正方形矩阵。

示例:

输入: 3
输出:
[
 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
]

来源：力扣（LeetCode）

链接：https://leetcode-cn.com/problems/spiral-matrix-ii

著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. LeetCode 59 解题

类似题目：LeetCode 885. 螺旋矩阵 III

创建变量top、bottom表示上下行的区间，left、right表示列的区间

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector<vector<int> > v(n, vector<int>(n,0));
        if(n == 0) 
        	return {};
        int i, num = 0, total = n*n, top = 0, bottom = n-1, left = 0, right = n-1;
        while(num < total)
        {
        	for(i = left; i <= right; ++i)
        		v[top][i] = ++num;
        	++top;
        	for(i = top; i <= bottom; ++i)
        		v[i][right] = ++num;
        	--right;
        	for(i = right; i >= left; --i)
        		v[bottom][i] = ++num;
        	--bottom;
        	for(i = bottom; i >= top; --i)
        		v[i][left] = ++num;
        	++left;
        }
        return v;
    }
};

3. LeetCode 54. 螺旋矩阵

给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        if(matrix.empty()) 
        	return {};
        vector<int> ans;
        int m = matrix.size(), n = matrix[0].size();
        int i, num = 0, total = m*n, top = 0, bottom = m-1, left = 0, right = n-1;
        while(num < total)
        {
        	for(i = left; i <= right; ++i)
        		ans.push_back(matrix[top][i]),num++;
            if(num == total)
                break;
        	++top;
        	for(i = top; i <= bottom; ++i)
        		ans.push_back(matrix[i][right]),num++;
            if(num == total)
                break;
        	--right;
        	for(i = right; i >= left; --i)
        		ans.push_back(matrix[bottom][i]),num++;
            if(num == total)
                break;
        	--bottom;
        	for(i = bottom; i >= top; --i)
        		ans.push_back(matrix[i][left]),num++;
            if(num == total)
                break;
        	++left;
        }
        return ans;
    }
};

4.《剑指Offer》面试题29

面试题29. 顺时针打印矩阵

class Solution {	//2020.2.22
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        if(matrix.size()==0 || matrix[0].size()==0)
        	return {};
        int i = 0, k = 0, count = matrix.size()*matrix[0].size();
        int left = 0, right = matrix[0].size()-1, up = 0, bottom = matrix.size()-1;
        vector<int> ans(count);
        while(count)
        {
        	i = left;
        	while(count && i <= right)
        	{
        		ans[k++] = matrix[up][i++];
        		count--;
        	}
        	up++,
        	i = up;
        	while(count && i <= bottom)
        	{
        		ans[k++] = matrix[i++][right];
        		count--;
        	}
        	right--;
        	i = right;
        	while(count && i >= left)
        	{
        		ans[k++] = matrix[bottom][i--];
        		count--;
        	}
        	bottom--;
        	i = bottom;
        	while(count && i >= up)
        	{
        		ans[k++] = matrix[i--][left];
        		count--;
        	}
            left++;
        }
        return ans;
    }
};