这本来不是要作为答案的，但实际上已经演变成了一个答案。老实说，我认为从Java开始并转向C是一个坏主意，因为这两种语言确实没有什么相似之处，而且您不会帮上忙，因为如果您依赖于Java的任何功能，在移植它时都会遇到严重的问题。有C没有（即大多数）
也就是说，我将草拟一些算法C的东西。
支持结构
typedefstruct Node{ int x, y; // x and y are array indices}Node;typedefstruct Path{ int maxlen, head; Node * path; // maxlen is size of path, head is the index of the current node // path is the pointer to the node array}Path;int node_compare(Node * n1, Node * n2); // returns true if nodes are equal, else falsevoid path_setup(Path * p, Node * n); // allocates Path.path and sets first nodevoid path_embiggen(Path * p); // use realloc to make path bigger in case it fills upint path_toosmall(Path * p); // returns true if the path needs to be reallocated to add more nodesNode * path_head(Path * p); // returns the head node of the pathvoid path_push(Path * p, Node * n); // pushes a new head node onto the pathvoid path_pop(Path * p); // pops a node from path您可能将迷宫格式更改为邻接列表之类的东西。您可以将每个节点存储为掩码，详细说明可以从该节点访问的节点。
迷宫格式
const int // these constants indicate which directions of travel are possible from a nodeN = (1 << 0), // travel NORTH from node is possibleS = (1 << 1), // travel SOUTH from node is possibleE = (1 << 2), // travel EAST from node is possibleW = (1 << 3), // travel WEST from node is possibleNUM_DIRECTIONS = 4; // number of directions (might not be 4. no reason it has to be)const intSTART = (1 << 4), // starting nodeFINISH = (1 << 5); // finishing nodeconst intMAZE_X = 4, // maze dimensionsMAZE_Y = 4;int maze[MAZE_X][MAZE_Y] = { {E, S|E|W, S|E|W, S|W }, {S|FINISH, N|S, N|START, N|S }, {N|S, N|E, S|E|W, N|S|W }, {N|E, E|W, N|W, N }};Node start = {1, 2}; // position of start nodeNode finish = {1, 0}; // position of end node我的迷宫与您的迷宫不同：两种格式彼此之间并非完全一对一地映射。例如，您的格式允许更精细的移动，但是我的格式允许单向路径。
请注意，您的格式明确指定了墙的位置。按照我的格式，墙在概念上位于无法通行的任何地方。我创建的迷宫有3个水平墙和5个垂直墙（并且也是封闭的，即有连续的墙围绕着整个迷宫）
为了进行遍历，我将使用深度优先搜索。您可以通过多种方式将标志映射到方向，例如以下所示。由于无论如何都要遍历每个对象，因此访问时间是无关紧要的，因此仅使用数组而不是某种更快的关联容器就足够了。
数据格式到偏移映射
// map directions to array offsets// format is [flag], [x offset], [y offset]int mappings[][] ={ {N, -1, 0}, {S, 1, 0}, {E, 0, 1}, {W, 0, -1}}最后，您的搜索。您可以迭代或递归实现。我的示例使用递归。
搜索算法伪代码
int search_for_path(int ** maze, char ** visited, Path * path){ Node * head = path_head(path); Node temp; int i; if (node_compare(head, &finish)) return 1; // found finish if (visited[head->x][head->y]) return 0; // don't traverse again, that's pointless visited[head->x][head->y] = 1; if (path_toosmall(path)) path_embiggen(path); for (i = 0; i < NUM_DIRECTIONS; ++i) { if (maze[head->x][head->y] & mappings[i][0]) // path in this direction { temp = {head->x + mappings[i][1], head->y + mappings[i][2]}; path_push(path, &temp); if (search_for_path(maze, visited, path)) return 1; // something found end path_pop(path); } } return 0; // unable to find path from any unvisited neighbor}要调用此功能，您应该像这样设置所有内容：
调用解算器
// we already have the maze// int maze[MAZE_X][MAZE_Y] = {...};// make a visited list, set to all 0 (unvisited)int visited[MAZE_X][MAZE_Y] = { {0,0,0,0}, {0,0,0,0}, {0,0,0,0}, {0,0,0,0}};// setup the pathPath p;path_setup(&p, &start);if (search_for_path(maze, visited, &path)){ // succeeded, path contains the list of nodes containing coordinates from start to end}else{ // maze was impossible}值得注意的是，因为我在编辑框中都写了这些，所以我还没有测试过。第一次尝试可能无法正常工作，可能需要花些时间。例如，除非在全局范围内声明了开始和结束，否则会有一些问题。最好将目标节点传递给搜索功能，而不要使用全局变量。
