54.螺旋矩阵
59.螺旋矩阵 II

54.螺旋矩阵

总体的思路分析：
顺时针，先遍历右边，再下面，再往左，再向上，然后再缩小一圈范围即可

原本的代码情况

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        m = len(matrix)
        n = len(matrix[0])
        ans = []
        left ,top = 0,0
        right,bottom = n-1,m-1
        while top <= bottom and left<= right:
            for i in range(left,right+1):
                ans.append(matrix[top][i])
            for i in range(top+1,bottom+1):
                ans.append(matrix[i][right])
            for i in range(right-1,left-1,-1):
                ans.append(matrix[bottom][i])
            for i in range(bottom-1,top,-1):
                ans.append(matrix[i][left])
            left,top = left+1,top+1
            right,bottom = right-1,bottom-1
        return ans

就是发现当面对长方形的时候会出现重复的元素

分析问题的错因？

再看另一个正确的代码

两个代码的区别就是对于边界值的处理，处理完边界值之后需要对边界值进行判断

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        m = len(matrix)
        n = len(matrix[0])
        ans = []
        left ,top = 0,0
        right,bottom = n-1,m-1
        while top <= bottom and left<= right:
            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
            if left>right:
                break

        return ans



        

59.螺旋矩阵II

这个是正方形，相对来说简单一点

我们可以考虑在完成一圈之后再对边界值修改

class Solution:
    def generateMatrix(self, n: int) -> List[List[int]]:
        left = 0
        right = n-1
        top = 0
        bottom = n-1
        matrix = [[0] * n for _ in range (n)]
        count = 1
        while top<=bottom and left<=right:
            for i in range(left,right+1):
                matrix[top][i] = count
                count+=1
            for i in range(top+1,bottom+1):
                matrix[i][right] = count
                count+=1
            for i in range (right-1,left-1,-1):
                matrix[bottom][i] = count
                count +=1
            for i in range (bottom-1,top,-1):
                matrix[i][left] = count
                count +=1
            top+=1
            right-=1
            bottom-=1
            left+=1
        return matrix        
        

为了应对长方形的情况，需要在每次修改边界之后对边界判断

class Solution:
    def generateMatrix(self, n: int) -> List[List[int]]:
        left = 0
        right = n-1
        top = 0
        bottom = n-1
        matrix = [[0] * n for _ in range (n)]
        count = 1
        while top<=bottom and left<=right:
            for i in range(left,right+1):
                matrix[top][i] = count
                count+=1
            top+=1
            if top > bottom:
                break

            for i in range(top,bottom+1):
                matrix[i][right] = count
                count+=1
            right-=1
            if right<left:
                break
            
            for i in range (right,left-1,-1):
                matrix[bottom][i] = count
                count +=1
            bottom-=1
            if bottom<top:
                break

            for i in range (bottom,top-1,-1):
                matrix[i][left] = count
                count +=1
            left+=1
            if left>right:
                break

        return matrix