LeetCode-探索-初级算法-字符串-2. 整数反转(个人做题记录,不是习题讲解)

LeetCode-探索-初级算法-字符串-2. 整数反转(个人做题记录,不是习题讲解)

LeetCode探索-初级算法:https://leetcode-cn.com/explore/interview/card/top-interview-questions-easy/

  1. 整数反转
  • 语言:java

  • 思路:判断正负号后,看是从0还是1下标截取字符串,反转后转换成int,然后根据正负添加符号,用try-catch处理溢出

  • 代码(3ms):

    class Solution {
        public int reverse(int x) {
            String num = String.valueOf(x);
            int start = 0;
            int end = num.length();
            if(x<0){
                start = 1;
            }
            StringBuilder str = new StringBuilder(num.substring(start,end));
            str.reverse();
            try{
                Integer res = Integer.parseInt(str.toString());
                if(start>0){
                    res = - res;
                }
                return res;
            }catch(Exception error){
                return 0;
            }
        }
    }
    
  • 参考代码(1ms):在知道最大和最小值个位是7个8的情况下,使用逐位*10加到结果ans的方法。

    class Solution {
        public int reverse(int x) {
            int ans = 0;
            while(x!=0)
            {
                int dig = x%10;
                if(ans > Integer.MAX_VALUE/10||(ans ==Integer.MAX_VALUE/10&&dig>7))
                {
                    return 0;
                }
                if(ans < Integer.MIN_VALUE/10||(ans ==Integer.MIN_VALUE/10&&dig<-8))
                {
                    return 0;
                }
                ans = ans*10 + dig;
                x = x /10;
            }
            return ans;
        }
    }
    
  • 参考后重写(1ms):

    class Solution {
        public int reverse(int x) {
            int res = 0;
            while(x!=0){
                int digit = x%10;
                if(res > Integer.MAX_VALUE/10||res ==Integer.MAX_VALUE/10&&digit>7)
                    return 0;
                if(res < Integer.MIN_VALUE/10||res ==Integer.MIN_VALUE/10&&digit<-8)
                    return 0;
                res = res * 10 + digit;
                x /= 10;
            }
            return res;
        }
    }
    

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