leetcode--矩阵遍历与搜索

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原创最新推荐文章于 2026-01-16 22:49:14 发布 437 阅读 AI 写同款· GEO 优化 ›

电流采样电路设计：差分放大与ADC采集详解在电机控制、电源管理、电池监测等应用中，电流采样是一个关键环节。通常，我们使用小阻值的采样电阻将电流信号转换为电压信号，再通过运放放大后送入ADC进行采集。本文将详细分析一种基于差分放大器的电流采样电路，包括信号放大、参考电压偏置、滤波保护以及ADC转换计算等内容。阅读详情

54. 螺旋矩阵

class Solution(object):
    def spiralOrder(self, matrix):
        res = []
        row, col= len(matrix)-1, len(matrix[0])-1
        i, j = 0, 0

        while i<=row and j <=col:
            if i == row:
                for x in range(j, col+1):
                    res.append(matrix[row][x])
                return res
            
            if j == col:
                for x in range(i, row+1):
                    res.append(matrix[x][col])
                return res

            for x in range(j, col+1):
                res.append(matrix[i][x])
            
            for x in range(i+1, row+1):
                res.append(matrix[x][col])


            for x in range(col-1, j-1, -1):
                res.append(matrix[row][x])
            
            for x in range(row-1, i, -1):
                res.append(matrix[x][j])
            i += 1
            j += 1
            row -= 1
            col -= 1
        return res

59. 螺旋矩阵 II

class Solution(object):
    def generateMatrix(self, n):
        """
        :type n: int
        :rtype: List[List[int]]
        """
        matrix = [[0] * n for _ in range(n)]
        i, j = 0, 0
        a, b=  n -1, n-1
        number = 1
        res = []
        while i <= a and j <= b:
            if i == a and j == b:
                matrix[i][j] = number
                return matrix
            
            for k in range(j, b+1):
                matrix[i][k] = number
                number += 1

            for k in range(i+1, a+1):
                matrix[k][b]  = number
                number += 1

            for k in range(b-1, j-1, -1):
                matrix[a][k] = number
                number += 1


            for k in range(a-1, i, -1):
                matrix[k][j] = number
                number += 1

            i += 1
            j += 1
            a -= 1
            b -= 1 

        return matrix

74. 搜索二维矩阵
 240. 搜索二维矩阵 II

class Solution(object):
    def searchMatrix(self, matrix, target):
        m,n = len(matrix)-1, len(matrix[0])-1
        
        i, j = m, 0

        while i>=0 and j<=n:
            if matrix[i][j] == target:
                return True
            elif matrix[i][j] > target:
                i -= 1
            else:
                j += 1
        
        return False