这是我的任务：编写一个程序，创建一个代表8×8网格的字符串，使用换行符分隔行。在网格的每个位置都有一个空格或#字符。角色应该成为棋盘。
将此字符串传递给console.log应该显示如下内容：
# # # #
# # # #
# # # #
# # # #
# # # #
# # # #
# # # #
# # # #
我的尝试低于评论以解释我的逻辑。我接近但我永远无法得到上面的模式。我的逻辑怎么了？
//declar empty board
var board = "";
//use outter loop to control rows
for (var y = 1; y < 5; y++) {
//use innerloop to generate the #'s horizonatlly
for(var x = 1; x < 5; x++) {
//if the row is even then put a space in front
//of the board string and then generate the board
// with the #'s
if (y % 2 === 0) {
board = " " + board;
board += "#";
}
//else generate the board without a space in front
//of the #'s
else {
board += "#";
}
}
//generate a new line
board += "\n";
}
console.log(board);
答案 0 :(得分：3)
我决定玩得开心并编写自己的解决方案：
var board = "";
var evenRow = "# # # # ";
var oddRow = " # # # #";
for ( var i = 0; i < 8; i++ ) {
if ( i%2 == 0 ) board += evenRow + '\n';
else board += oddRow + '\n';
}
console.log(board);
注意：国际象棋棋盘的正确方向是右下方的白色方块，这是答案的建议输出的90度旋转。
答案 1 :(得分：0)
上面的解决方案很适合用于循环。让循环播放很有趣。
var board = "";
var even = " # # # #";
var odd = "# # # #";
var i = 1;
while (i <= 8) {
if (i % 2 == 0) {
console.log(even);
} else {
console.log(odd);
}
var i = i + 1;
}
答案 2 :(得分：0)
解决方案：
Pull request creation failed. Validation failed: A pull request already exists for...答案 3 :(得分：0)
这是我的解决方案，可以产生所需的准确输出
// declare size and an empty "board" string
var size = 8;
var board = "";
// Outerloop for rows
for (let i = 1; i <= size; i++) {
// inner loop for columns
for (let j = 1; j <= size; j++) {
if ((i + j) % 2 == 0) { // check if col is even
board += " ";
} else { // if col is odd
board += "#";
}
}
board += "\n"; // jump to next row
}
console.log(board);
答案 4 :(得分：0)
let size = 8;
let board = "";
for (let y = 0; y < size; y++) {
for (let x = 0; x < size; x++) {
if ((x + y) % 2 == 0) {
board += " ";
} else {
board += "#";
}
}
board += "\n";
}
console.log(board);
答案 5 :(得分：0)
我想出了这个实现，我根据所需的网格大小使用循环创建所需的字符串行，最后使用另一个循环将它们与新行连接以输出最终结果。
//I Came Up With This
const gridSize = 8 ; //your grid size
var str1 = "" ; //first line of string eg. # # # #
var str2 = "" ; //second line of string eg.# # # #
var str = "" ; //final string
//making str1 and str2 according to gridSize
for(var i = 0 ; i < gridSize ; i++){
if(i%2==0){
str1 += " " ;
str2 += "#" ;
}
else{
str1 += "#" ;
str2 += " " ;
}
}
//concatenating str1 and str2 with new line to final output str.
for(var i = 0 ; i < gridSize ; i++){
if(i%2==0){
str += str1+"\n" ;
}
else{
str += str2+"\n" ;
}
}
//printing final output.
console.log(str);答案 6 :(得分：-1)
这是我的解决方案：
var size = 8;
var block = '#';
var space = ' ';
for (var i = 1; i <= size; i++) {
var line = '';
for (var y = 1; y <= size; y++) {
if (i % 2) {
if (y % 2) {
line = line + space;
} else {
line = line + block;
}
} else {
if (y % 2) {
line = line + block;
} else {
line = line + space;
}
}
}
console.log(line);
}
答案 7 :(得分：-1)
var x = "";
var y = "";
for (var i = 0; i <= 4; i++) {
x = "#" + " " + "#" + " " + "#" + " " + "#";
y = " " + "#" + " " + "#" + " " + "#" + " " + "#";
console.log(x);
console.log(y);
}
答案 8 :(得分：-1)
这是我的解决方案......
var datepickerCalendarDirective = function datepickerCalendarDirective() {
return {
scope: {
searchResponse: '='
},
restrict: 'E',
templateUrl: './templates/directive.datepickerCalendar.tmpl.html',
link: function link(scope, element, attrs, $filter, policySessionService) {
scope.$watch('searchResponse', function () {
scope.dt = new Date();
if (!scope.searchResponse) return '';
/* setup the date picker options object.
* this takes a customClass function which sets the
* css class of the trips/events on the datepicker
*/
scope.options = {
customClass: getSelectedDates,
maxDate: new Date(),
showWeeks: false,
startingDay: 1 };
}, true);
/* Iterate over each trip in the reponse and return either
* a trip or incident css class to the customClass param
* of the datepicker options object
*/
function getSelectedDates(data) {
if (scope.searchResponse.data[0] && data.mode === 'day') {
var trips = $filter('orderBy')(scope.searchResponse.data[0].tripDates, 'JourneyDate'),
journeyDate;
for (var i = 0; i < trips.length; i++) {
journeyDate = new Date(trips[i].JourneyDate).setHours(0, 0, 0, 0);
if (journeyDate === new Date(data.date).setHours(0, 0, 0, 0)) {
// return the appropriate css class for the trip type (trip or incident)
if (trips[i].IncidentRecorded) {
return 'incident';
} else {
return 'trip';
}
}
}
}
return '';
}
// });
}
};
};
答案 9 :(得分：-2)
这个实现如何简单直接
function chessBoard(size) {
for(var i = 0; i < size; i++) {
if (i % 2 === 0) {
console.log('' + '# # # #');
} else {
console.log(' ' + '# # # #');
}
}
}
chessBoard(8);
答案 10 :(得分：-3)
这很简单...... 2个循环...第一个循环计算列数，第二个循环每列重复“＃”4次。与“＃”的距离是字符“＃”之间的距离。当列数为奇数时，距离将在“＃”之前。变量“chr”是将这些重复内容保存到每一列中。
<select class="form-control input-sm" ng-model="selectedDay" ng-options="day.id as day.name for day in daysOfWeek" ng-change="triggerDate(dt)"></select>