Question

我已经编写了一个递归实现，用于使用我在stackoverflow上找到的psuedocode给出一些n * n矩阵来查找某个特定路径。我构建了两个数组：

路径权重数组，用于保存从图形中的一个顶点到另一个顶点的边缘值（单元格矩阵到矩阵中的相邻单元格）
路径权重数组中每个索引的邻接数组

出于某种原因，但是当我的递归函数返回时，它不会返回到直接调用者？

我的实施：

        // find paths of specific length from our starting cell 1,1
        // when a path is found print it
        // otherwise return? 
        function findPaths(adjMatrix, pathW_array, path, pathWeight, x, y, idx, curpath_idx){
            // set curpath to the current path (an array with storing path weight values)
            var curpath = path; 
            // tempArray with store pathW_array (set cell value to -1 if value has been added to curpath)
            var tempArray = pathW_array; 

            // curValue retruns sum value of our curpath
            if ( (curValue(curpath) == pathWeight) && (Object.keys(curpath).length !=1)){
                for(i = 0; i < Object.keys(curpath).length; i++){
                    console.log(curpath[i]);
                }
                return;
            }if(tempArray[x][y] == -1){
                return;
            }if(curValue(curpath) > pathWeight){
                return;
            }   

            // Did not return, add current cell value to curpath array
            curpath[curpath_idx] = tempArray[x][y]; 
            curpath_idx = curpath_idx + 1;
            // set current cell value in tempArray to -1 because we've added it to current path (do not want to add same cell value multiple times)
            tempArray[x][y] = -1;

            // iterate until i = pathWeight -1 
            for(var i = 0; i < pathWeight; i++){
                if(adjMatrix[idx][i] != -1){
                     // get pathW_array indices for next neighbor cell of current element
                     arrayIndices = neighbor_value(adjMatrix, tempArray, idx);
                     x = arrayIndices[0];
                     y = arrayIndices[1]; 
                     adjMatrix[idx][i] = -1;
                     idx = adjMatrix_idx(x,y);

                    // findpaths from the next cell in the matrix 
                    findPaths(adjMatrix, tempArray, curpath, pathWeight, x, y, idx, curpath_idx);
                    curpath.pop();
                    curpath_idx = curpath -1;
                }   
            }
        }

Answer 1

更改以下行：

findPaths(adjMatrix, tempArray, curpath, pathWeight, x, y, idx, curpath_idx)

到

return findPaths(adjMatrix, tempArray, curpath, pathWeight, x, y, idx, curpath_idx)

如果不理解函数中其余的复杂性，那么'应该'将递归返回值返回到调用堆栈。

递归函数不返回直接调用者

1 个答案: