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Find an integer between 1 and 9999, determine whether it is a palindrome number, if so, output and output the number of digits in this number
<script type="text/javascript">/*先整理思路、、、、、 var n = 3223; 只要前后颠倒相等就是回文数 整数的位数: n.length for(var i = n.length - 1; i <= o ; i--){ m.push(n[i]); } document.write(m); */ function huiwen(){ var num = prompt("请输入一个整数"); document.write("你输入的数是:" + num + "<br>"); var n =""; //定义颠倒后的字符串 if(num<1){ document.write("不要输入负数"); }else if (num >= 1 &&num <10){ document.write("这是一个回文数" + "<br>"); document.write("这个回文数的位数是:1"); } else{ for(var i = num.length ; i >= 0 ; i--){ var num1 = num.charAt(i); //charAt等同于数组的下标,这样通过for循环就可以把输入的整数顺序颠倒 n = n.concat(num1); // 但是num1并不是一个整数,也不是一个字符串,没法跟num比较,用concat()将它们连接成一个字符串 } document.write( n + ":"); //得到的回文数输出一下 if( n == num){ document.write("这是一个回文数" + "<br>"); document.write("这个回文数的位数是:" + n.length) }else{ document.write('这不是一个回文数'); } } }</script> </head> <body> <input type="button" value="开始" onClick="huiwen()"> </body>
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