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lintcode-easy-Reverse Integer

时间:2016-03-06 11:24:46      阅读:133      评论:0      收藏:0      [点我收藏+]

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Reverse digits of an integer. Returns 0 when the reversed integer overflows (signed 32-bit integer).

 

Given x = 123, return 321

Given x = -123, return -321

这题主要注意一下溢出的可能

public class Solution {
    /**
     * @param n the integer to be reversed
     * @return the reversed integer
     */
    public int reverseInteger(int n) {
        // Write your code here
        long result = 0;
        long buf = n;
        boolean neg = false;
        if(n < 0){
            neg = true;
            buf = -n;
        }
        
        while(buf > 0){
            result = 10 * result + buf % 10;
            buf /= 10;
        }
        
        if(neg)
            result = -result;
        
        if(result > Integer.MAX_VALUE || result < Integer.MIN_VALUE)
            return 0;
        
        return (int)result;
        
    }
}

 

lintcode-easy-Reverse Integer

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原文地址:http://www.cnblogs.com/goblinengineer/p/5246651.html

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