Le problème des billets de voyage

Ce qui suit est un échantillon de la mise en œuvre JavaScript.

var un  =
[
 { from:'c',to:'d'},
 { from:'a',to:'b'},
 { from:'b',to:'c'},
 { from:'z',to:'a'},
 { from:'d',to:'m'},
]

function buildTable( un ){
return un.reduce(function(previousValue, currentValue, currentIndex ){

//build the table.
    previousValue.from[currentValue['from']] = currentValue;
    previousValue.to[currentValue['to']] = currentValue;

//to find start and end.    
    if( !previousValue.from[currentValue['to']] )  previousValue.from[currentValue['to']]= false;
    if(!previousValue.to[currentValue['from']])  previousValue.to[currentValue['from']]= false;

    return previousValue;

},{to:{},from:{}} );


}

function getStart(nodes){
//find start node indx.
    for(var i  in nodes)
    if( !nodes[i] )return i;
}


function print(start,nodes ){


var sorted = [];
//while detecting false from buildTable structure.
    while(start){
        var  node = nodes[start];
        sorted.push(node)
        start = node['to'];
//console.log(start)
    }

return sorted;
}

var sorted = buildTable(un);
console.log(print(getStart(sorted.to),sorted.from));

    function fullPath(path) {
    let startingPoints = [];
    let endingPoints = [];
    let result = [];

    for (let i = 0; i < path.length; i++) {
        startingPoints.push(path[i].from);
        endingPoints.push(path[i].to);
    }

    for (let i = 0; i < path.length; i++) {
        if (!endingPoints.includes(path[i].from)) {
            result.push(path[i].from);
            break;
        }
    }

    for (let i = 0; i < path.length; i++) {
        if (startingPoints.includes(result[i])) {
            let indexOfPoint = startingPoints.indexOf(result[i]);
            result.push(endingPoints[indexOfPoint]);
        }
    }
    return result;

}