LeetCode-7 Reverse Integer(倒置整型数)

发布时间：2020-12-13 22:29:00 所属栏目：百科来源：网络整理

导读：LeetCode-7 Reverse Integer Reverse digits of an integer. Example1: x = 123,return 321 Example2: x = -123,return -321 click to show spoilers. Have you thought about this? Here are some good questions to ask before coding. Bonus points for y

LeetCode-7 Reverse Integer

Reverse digits of an integer.

Example1:x = 123,return 321
Example2:x = -123,return -321

click to show spoilers.

Have you thought about this?

Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer's last digit is 0,what should the output be? ie,cases such as 10,100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer,then the reverse of 1000000003 overflows. How should you handle such cases?

For the purpose of this problem,assume that your function returns 0 when the reversed integer overflows.

Update (2014-11-10):
Test cases had been added to test the overflow behavior.

直接string倒置，

要留意NumberFormatException的问题，

通过catch这个异常来解决倒置后溢出的问题。

public class Solution {
	public int reverse(int x) {
		String prefix = ""; // 用作保存前缀
		if (x < 0) {
			x = -x;
			prefix = "-";
		}
		StringBuilder sb = new StringBuilder(prefix);
		String reverse = new StringBuilder(String.valueOf(x)).reverse()
				.toString();
		String str = sb.append(reverse).toString();

		try {
			return Integer.parseInt(str);
		} catch (NumberFormatException e) {
			return 0;
		}
	}
}

Runtime:275 ms

关于看到的另一种做法：

public class Solution {
    public int reverse(int x) {
        int output = 0;
    	for(;x!=0;x/=10){
    		output=output*10+x%10;
    	}
    	return output;
    }
}

这种做法觉得很强大代码很简洁~这是没考虑溢出的代码，

但加入溢出情况的考虑后代码会比较多~不知道有没有其他同学有更简洁的写法~

如下：

public class Solution {
	public int reverse(int x) {
		boolean ifnegative = false;// 是否为负数
		if (x < 0) {
			// integer的范围是-2147483648~2147483647，
			// 取绝对值会漏了-2147483648的情况，所以在这里单独考虑。
			if (x == -2147483648) {
				return 0;
			} else {
				x = Math.abs(x);
				ifnegative = true;// 标记为负数
			}
		}
		int output = 0;
		for (; x != 0; x /= 10) {
			output = output * 10 + x % 10;
			if (output >= 214748364 && x >= 10) {
				if (output == 214748364) {// 若个位数以前未溢出
					if ((x / 10) > 7) {// 判断个位数是否会导致溢出
						return 0;
					} else {
						continue;
					}
				} else {// 即 output>214748364的情况，十位上已经溢出。
					return 0;
				}
			}
		}
		if (ifnegative) {
			return -output;
		} else {
			return output;
		}
	}
}

Runtime:253 ms

（编辑：李大同）

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