leetcode Spiral Matrix || leetcode Spiral Matrix II

leetcode Spiral Matrix

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].

题目地址：leetcode Spiral Matrix

题意：给定一个矩阵，按题目要求求出序列。。。

思路：直接循环模拟即可

C++

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

            for(int i = top; i <= bottom; ++i)
                 ans.push_back(matrix[i][right]);
            if(--right < left) break;
            
            for(int i = right; i >= left; --i)
                 ans.push_back(matrix[bottom][i]);
            if(--bottom < top) break;
            
            for(int i = bottom; i >= top; --i)
                 ans.push_back(matrix[i][left]);
            ++left;
        }
        return ans;
    }
};

Python

class Solution(object):
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        ans = []
        if not matrix: return ans
        top, bottom, left, right = 0, len(matrix) - 1, 0, len(matrix[0]) - 1
        while left <= right and top <= bottom:
            for i in range(left, right + 1):
                ans.append(matrix[top][i])
            top += 1
            if top > bottom: break
            
            for i in range(top, bottom + 1):
                ans.append(matrix[i][right])
            right -= 1
            if right < left: break
            
            for i in range(right, left - 1, -1):
                ans.append(matrix[bottom][i])
            bottom -= 1
            if bottom < top: break
            
            for i in range(bottom, top - 1, -1):
                ans.append(matrix[i][left])
            left += 1
            
        return ans

 

leetcode Spiral Matrix II

Given an integer n, generate a square matrix filled with elements from 1 to _n_2 in spiral order.

For example, Given n = 3,

You should return the following matrix:

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

题目地址：leetcode Spiral Matrix II

题意:这题和上面一题的区别在于，这题是给n，求从1~n组从的序列

思路直接用上面的代码改改即可。

C++

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector<vector<int>> ans(n, vector<int>(n));
        int left = 0, right = n - 1, top = 0, bottom = n - 1;
        int cur = 0;
        while(left <= right && top <= bottom){
            for(int i = left; i <= right; ++i)
                ans[top][i] = ++cur;
            if(++top > bottom) break;

            for(int i = top; i <= bottom; ++i)
                ans[i][right] = ++cur;
            if(--right < left) break;
            
            for(int i = right; i >= left; --i)
                ans[bottom][i] = ++cur;
            if(--bottom < top) break;
            
            for(int i = bottom; i >= top; --i)
                ans[i][left] = ++cur;
            ++left;
        }
        return ans;
    }
};

Python

class Solution(object):
    def generateMatrix(self, n):
        """
        :type n: int
        :rtype: List[List[int]]
        """
        ans = [[0] * n for _ in range(n)]
        top, bottom, left, right = 0, n - 1, 0, n - 1
        cur = 1
        while left <= right and top <= bottom:
            for i in range(left, right + 1):
                ans[top][i] = cur
                cur += 1
            top += 1
            if top > bottom: break
            
            for i in range(top, bottom + 1):
                ans[i][right] = cur
                cur += 1
            right -= 1
            if right < left: break
            
            for i in range(right, left - 1, -1):
                ans[bottom][i] = cur
                cur += 1
            bottom -= 1
            if bottom < top: break
            
            for i in range(bottom, top - 1, -1):
                ans[i][left] = cur
                cur += 1
            left += 1
            
        return ans