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+ package com .blankj .easy ._007 ;
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+
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+ /**
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+ * @author luojing
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+ * @description LJSolution
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+ * @date 2022/07/15
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+ */
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+
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+ public class LJSolution {
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+ public int reverse (int x ) {
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+ long res = 0 ;
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+ for (; x != 0 ; x /= 10 ) {
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+ /**
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+ * 0. 除10一循环,直到除不了10后
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+ * 1. 先获取x的余数,也就是最后一位,依次类推获取新的最后一位
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+ * 2. 获取最后一位数后再按次数乘以10,就将整形数反转了
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+ * 3. 溢出返回0,返回时判断下是否会溢出
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+ * 4. 注意0*10还是0,不存在反转后001这种数
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+ */
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+ res = res * 10 + x %10 ;
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+ }
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+ return res > Integer .MAX_VALUE || res <Integer .MIN_VALUE ? 0 : (int ) res ;
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+ }
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+
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+ public static void main (String [] args ) {
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+ LJSolution ljSolution = new LJSolution ();
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+ System .out .println (ljSolution .reverse (123 ));
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+ System .out .println (ljSolution .reverse (-123 )); //负数
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+ System .out .println (ljSolution .reverse (100 )); //末尾有0
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+ System .out .println (ljSolution .reverse (1000000003 )); //溢出
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+ }
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+ }
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