Runtime: 5 ms, faster than 11.06% of Java online submissions for Reverse Integer.Memory Usage: 38.8 MB, less than 7.78% of Java online submissions for Reverse Integer.

abc = a * 10^2 + b * 10^1 + c * 10^0;

intpop=x%10;//余数x=x/10;//处的结果// 反过来的结果计算result=result*10+pop;

[pop]: 5, [x]: 1234, [result]: 5[pop]: 4, [x]: 123, [result]: 54[pop]: 3, [x]: 12, [result]: 543[pop]: 2, [x]: 1, [result]: 5432[pop]: 1, [x]: 0, [result]: 54321

publicintreverse(intx){longresult=0;while(x!=0){// 移除最后一位intpop=x%10;x=x/10;// 返回值result=result*10+pop;}// 越界判断if(result>Integer.MAX_VALUE||result<Integer.MIN_VALUE){return0;}return(int)result;}

Runtime: 1 ms, faster than 100.00% of Java online submissions for Reverse Integer.Memory Usage: 36.4 MB, less than 94.78% of Java online submissions for Reverse Integer.

输入：x = 121输出：true

输入：x = -121输出：false解释：从左向右读, 为 -121 。 从右向左读, 为 121- 。因此它不是一个回文数。

输入：x = 10输出：false解释：从右向左读, 为 01 。因此它不是一个回文数。``` ### 提示：-2^31 <= x <= 2^31 - 1### 进阶：你能不将整数转为字符串来解决这个问题吗？## V1-转换为字符串处理### 思路我们可以把 int 转换为 string，然后使用前面对于字符串回文判断方式，来解决。### java 实现```javapublic boolean isPalindrome2(int x) {    String string = String.valueOf(x);    final int length = string.length();    int mid = length >> 1;    // 从中间往两边均摊    for (int i = 0; i < mid; i++) {        if (string.charAt(i) != string.charAt(length - i - 1)) {            return false;        }    }    return true;}

Runtime: 7 ms, faster than 76.19% of Java online submissions for Palindrome Number.Memory Usage: 38.5 MB, less than 92.10% of Java online submissions for Palindrome Number.

publicbooleanisPalindrome(intx){if(x<0){returnfalse;}// 反转intreverse=reverse(x);returnx==reverse;}privateintreverse(intx){intresult=0;while(x!=0){// 移除最后一位intpop=x%10;x=x/10;// 返回值result=result*10+pop;}returnresult;}

publicbooleanisPalindrome(intx){inttemp=x;intds=0;while(temp>0){// 直接反向操作，计算逆序的数值ds=ds*10+(temp%10);// 保留的高位temp/=10;}return(x-ds)==0;}

Runtime: 7 ms, faster than 96.19% of Java online submissions for Palindrome Number.Memory Usage: 39.3 MB, less than 86.45% of Java online submissions for Palindrome Number.

字符          数值I             1V             5X             10L             50C             100D             500M             1000

输入: num = 3输出: "III"

输入: num = 4输出: "IV"

输入: num = 9输出: "IX"

输入: num = 58输出: "LVIII"解释: L = 50, V = 5, III = 3.

输入: num = 1994输出: "MCMXCIV"解释: M = 1000, CM = 900, XC = 90, IV = 4.

/**     * 思路1：basic     * 思路2：HashMap     *     * 1 I     * 2 II     * 3 III     * 4 IV = 5-1     * 5 V = 5     * 6 VI = 5+1     * 7 VII     * 8 VIII     * 9 IX = 10-1     * 10 X = 10     *     * 考虑范围：1-3999     *     * 最多有 4 位     *     * 是否需要栈？     * @param num 数字     * @return 结果     * @since v1     */publicStringintToRoman(intnum){Stack<Integer>stack=newStack<>();while(num>0){intmod=num%10;// 剩余的部分num=num/10;// 余数stack.add(mod);}StringBuilderstringBuilder=newStringBuilder();while(!stack.isEmpty()){// 第一个出来的是最高位intnumber=stack.pop();// 跳过if(number==0){continue;}if(stack.size()==3){stringBuilder.append(build(number,"M","",""));}elseif(stack.size()==2){stringBuilder.append(build(number,"C","D","M"));}elseif(stack.size()==1){stringBuilder.append(build(number,"X","L","C"));}elseif(stack.size()==0){stringBuilder.append(build(number,"I","V","X"));}}returnstringBuilder.toString();}/**     * 核心实现     * @param number 数字     * @param currentChar  当前字符     * @param fiveChar 中间     * @param tenChar 最后     * @return 结果     * @since v1     */privateStringbuild(finalintnumber,finalStringcurrentChar,finalStringfiveChar,finalStringtenChar){if(number<=3){returnrepeat(number,currentChar);}elseif(number==4){returncurrentChar+fiveChar;}elseif(number==5){returnfiveChar;}elseif(number==9){returncurrentChar+tenChar;}else{returnfiveChar+repeat(number-5,currentChar);}}/**     * 重复构建多次     * @param times 次数     * @param c 字符     * @return 结果     * @since v1     */privateStringrepeat(finalinttimes,finalStringc){StringBuilderstringBuilder=newStringBuilder();for(inti=0;i<times;i++){stringBuilder.append(c);}returnstringBuilder.toString();}

Runtime: 14 ms, faster than 27.42% of Java online submissions for Palindrome Number.Memory Usage: 44.3 MB, less than 23.25% of Java online submissions for Palindrome Number.

/**     * 利用减法替代除法+乘法     *     * @param num 数字     * @return 结果     * @since best     */publicStringintToRoman(intnum){StringBuildersb=newStringBuilder();while(num>0){if(num>=1000){num-=1000;sb.append("M");}elseif(num>=900){num-=900;sb.append("CM");}elseif(num>=500){num-=500;sb.append("D");}elseif(num>=400){num-=400;sb.append("CD");}elseif(num>=100){num-=100;sb.append("C");}elseif(num>=90){num-=90;sb.append("XC");}elseif(num>=50){num-=50;sb.append("L");}elseif(num>=40){num-=40;sb.append("XL");}elseif(num>=10){num-=10;sb.append("X");}elseif(num>=9){num-=9;sb.append("IX");}elseif(num>=5){num-=5;sb.append("V");}elseif(num>=4){num-=4;sb.append("IV");}else{num-=1;sb.append("I");}}returnsb.toString();}

classSolution{int[]nums=newint[]{1000,900,500,400,100,90,50,40,10,9,5,4,1};String[]strings=newString[]{"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};/**     * 利用减法替代除法+乘法     *     * @param num 数字     * @return 结果     * @since best     */publicStringintToRoman(intnum){StringBuildersb=newStringBuilder();while(num>0){for(inti=0;i<nums.length;i++){intcurNum=nums[i];if(num>=curNum){num-=curNum;sb.append(strings[i]);break;}}}returnsb.toString();}}

Runtime: 4 ms, faster than 99.96% of Java online submissions for Palindrome Number.Memory Usage: 41.6 MB, less than 97.7% of Java online submissions for Palindrome Number.