根据json的from&to遍历出多少种组合

Question

javascriptjson算法

Answer 1

如果你的分支节点按先后次序出现，而不是加插到其他位置，可以这样遍历，后面有///的是新增来测试的点    var d = { "lines": {
        "demo_line_3": {
            "from": "demo_node_1",
            "to": "demo_node_2"
        },
        "demo_line_5": {
            "from": "demo_node_2",
            "to": "demo_node_4"
        },
        "demo_line_8": {
            "from": "demo_node_4",
            "to": "demo_node_6"
        },
        "demo_line_9": {
            "from": "demo_node_4",
            "to": "demo_node_7"
        },
        "demo_line_91": {
            "from": "demo_node_4",
            "to": "demo_node_71"///////
        },
        "demo_line_11": {
            "from": "demo_node_6",
            "to": "demo_node_10"
        },
        "demo_line_12": {
            "from": "demo_node_7",
            "to": "demo_node_10"
        },
        "demo_line_14": {
            "from": "demo_node_10",
            "to": "demo_node_13"
        },
        "demo_line_15": {
            "from": "demo_node_13",
            "to": "demo_node_15"////////
        }
    }
    };
    var paths = [],line,arr;
    for (attr in d.lines) {
        line = d.lines[attr];
        if (paths.length == 0) paths[0] = { path: line.from + '->' + line.to, to: line.to, from: line.from };
        else {
            for (var i = 0, l = paths.length; i < l; i++) {
                if (paths[i].to == line.from) { paths[i].path += '->' + line.to; paths[i].to = line.to; paths[i].from = line.from; }
                else if (paths[i].from == line.from) {//分裂线路
                    arr = paths[i].path.split('->');
                    arr[arr.length - 1] = line.to;
                    paths[paths.length] = { path: arr.join('->'), from: line.from, to: line.to };
                    break;
                }
            }
        }
    }    for (var i = 0; i < paths.length; i++)
        document.write(paths[i].path + '<br>');