给你一个 32 位的有符号整数 x ,返回 x 中每位上的数字反转后的结果。如果反转后整数超过 32 位的有符号整数的范围 [−231, 231 − 1] ,就返回 0。假设环境不允许存储 64 位整数(有符号或无符号)。示例 1:输入:x = 123输出:321示例 2:输入:x = -123输出:-321示例 3:输入:x = 120输出:21示例 4:输入:x = 0输出:0
function handle(x){
if(x<0){
var chooce = -1
x = -x;
var numLength = x.toString().length;
var sum = 0;
var discard = 0;
for(var i=0;i<numLength;i++){
var j = numLength-(i+1);
var num = Math.floor((x - discard) / 10**j);
discard = discard + num * 10**j;
sum = sum + num * 10**i
}
if(sum>2**31 || sum<-(2**31)){
return 0
}
return -sum
}else{
var chooce = 1
var numLength = x.toString().length;
var sum = 0;
var discard = 0;
for(var i=0;i<numLength;i++){
var j = numLength-(i+1);
var num = Math.floor((x - discard) / 10**j);
discard = discard + num * 10**j;
sum = sum + num * 10**i
}
console.log(sum)
console.log(2**32)
if(sum>2**31 || sum<-(2**31)){
return 0
}
return sum
}
}
function handle2(x){
var choose = (x>0)?'1':'-1'
x = x * choose
var numLength = x.toString().length;
var sum = 0;
var discard = 0;
for(var i=0;i<numLength;i++){
var j = numLength-(i+1);
var num = Math.floor((x - discard) / 10**j);
discard = discard + num * 10**j;
sum = sum + num * 10**i
}
console.log(sum)
console.log(2**32)
if(sum>2**31 || sum<-(2**31)){
return 0
}
return sum * choose
}
console.log(handle(1563847412));