【剑指offer】矩阵中的路径（回溯法）

分手后的思念是犯贱 2022-05-09 13:38 179阅读 0赞

![在这里插入图片描述][70]

# -*- coding:utf-8 -*-
    class Solution:
        def hasPathCore(self, matrix, rows, cols, row, col, path, pathIndex, visited):
            if pathIndex == len(path): return True
            #print(row, col)
            hasPath = False
            if row >= 0 and row < rows and col >= 0 and col < cols and matrix[row*cols+col] == path[pathIndex] and not visited[row][col]:
                visited[row][col] = 1
                #print('success', row, col)
                pathIndex += 1
                hasPath = self.hasPathCore(matrix, rows, cols, row, col - 1, path, pathIndex, visited) or \
                    self.hasPathCore(matrix, rows, cols, row, col + 1, path, pathIndex, visited) or \
                    self.hasPathCore(matrix, rows, cols, row + 1, col, path, pathIndex, visited) or \
                    self.hasPathCore(matrix, rows, cols, row - 1, col, path, pathIndex, visited)
                if not hasPath:
                    pathIndex -= 1  #回溯
                    visited[row][col] = 0
            return hasPath
        def hasPath(self, matrix, rows, cols, path):
            # write code here
            index = 0
            if len(matrix) == 0 or rows < 1 or cols < 1 or path == []:
                return False
            visite = [([0] * cols) for i in range(rows)]
            pathIndex = 0
            for i in range(cols):
                for j in range(rows):
                    if self.hasPathCore(matrix, rows, cols, j, i, path, pathIndex, visite):
                        return True
            return False
    matrix = 'ABCESFCSADEE'
    rows = 3
    cols = 4
    path = 'ABCB'
    print(Solution().hasPath(matrix, rows, cols, path))

采用回溯法来寻找路径，遍历四个方向，如果不符合，则掉头  
因此回溯法通常跟递归一起使用，这样方便返回上层

[70]: /images/20220506/d4c403bc28b143d2b14b5d0f4985cea3.png