android/移动端面试#1 大数相加

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最近在找工作，有一些面试，有一些面试题，做点整理和回顾吧。

题目是：

有两个大数（Int32放不下的），要求他们的和

一接手我就考虑用字符串的模拟加法来做，然后就和面试官说了，可能当时时间有点赶，面试官没要求我写代码出来。

今天自己计时写了一下，这里给下代码：

public String computer(String one,String another) throws IllegalArgumentException{if(one == null || another == null || one.isEmpty() || another.isEmpty()){throw new IllegalArgumentException("input is empty");}int olength = one.length();int alength = another.length();int minlength = Math.min(olength, alength);int length = Math.max(olength,alength) + 1;byte[] out = new byte[length];char zero = '0';int o,a;int current = 0 ;boolean flag = false;int k = length - 1;for(int i = olength - 1,j = alength - 1; i >= 0 && j >= 0 ; i--,j--){o = one.charAt(i) - zero;a = another.charAt(j) - zero;if(o < 0 || o > 9 || a < 0 || a > 9){throw new IllegalArgumentException("input not a num(0~9),check your input");}current = o + a ;if(flag) {current ++;}if(current > 9){flag = true;current -= 10; }else{flag = false;}out[k] = (byte)(zero + current);k--;}if(olength > minlength){flag = addLast(zero,one,flag,k,out);}else if(alength > minlength){flag = addLast(zero,another,flag,k,out);}out[0] = '1';if(flag){return new String(out);}else{return new String(out).substring(1);}}private boolean addLast(char zero,String s,boolean flag,int k,byte[] out)throws IllegalArgumentException{int t;int current;for(int i = k - 1 ; i >= 0 ; i-- ){t = s.charAt(i) -zero;if(t < 0 || t > 9){throw new IllegalArgumentException("input not a num(0~9),check your input");}current = t;if(flag){current++;}if(current > 9){flag = true;current -= 10; }else{flag = false;}out[k] = (byte)(zero + current);k--;}return flag;}

简单测试用例：

System.out.println("-----test1------");try{String a =    "789";String b ="90009999999";String c = computer(a, b);System.out.println(a);System.out.println(b);System.out.println(c);}catch(IllegalArgumentException e){e.printStackTrace();}System.out.println("-----test2------");try{String a = "1116666666666666666666611111188789";String b = "1";String c = computer(a, b);System.out.println(a);System.out.println(b);System.out.println(c);}catch(IllegalArgumentException e){e.printStackTrace();}System.out.println("-----test3------");try{String a =  "123456789";String b ="809999999";String c = computer(a, b);System.out.println(a);System.out.println(b);System.out.println(c);}catch(IllegalArgumentException e){e.printStackTrace();}System.out.println("-----test4------");try{String a =  "999";String b ="111";String c = computer(a, b);System.out.println(a);System.out.println(b);System.out.println(c);}catch(IllegalArgumentException e){e.printStackTrace();}System.out.println("-----test5------");try{String a =  "1234num56789";String b ="809999999";String c = computer(a, b);System.out.println(a);System.out.println(b);System.out.println(c);}catch(IllegalArgumentException e){e.printStackTrace();}

回头看网上的代码，找到一个代码行数大概在35行的，他对string做了反转，再用了一个int[]的数字去存结果，最后才转化过去。

然后我就发现可以直接用一行代码实现：

new BigInteger(one).add(new BigInteger(another)).toString();

（实践中确实很少遇见这种要求，所以，真不知道有这个库函数-_-）

顺便看看一下openjdk中是怎么写的吧：

看了下代码行数，有三千多行，我只挑重点的看吧。

1、先把字符串按照正负、进制（2~36）存在一个符号位和数据位int[]中

2、如果符号位相同（加法）分段求和

    /**     * Adds the contents of the int arrays x and y. This method allocates     * a new int array to hold the answer and returns a reference to that     * array.     */    private static int[] add(int[] x, int[] y) {        // If x is shorter, swap the two arrays        if (x.length < y.length) {            int[] tmp = x;            x = y;            y = tmp;        }        int xIndex = x.length;        int yIndex = y.length;        int result[] = new int[xIndex];        long sum = 0;        // Add common parts of both numbers        while(yIndex > 0) {            sum = (x[--xIndex] & LONG_MASK) +                  (y[--yIndex] & LONG_MASK) + (sum >>> 32);            result[xIndex] = (int)sum;        }        // Copy remainder of longer number while carry propagation is required        boolean carry = (sum >>> 32 != 0);        while (xIndex > 0 && carry)            carry = ((result[--xIndex] = x[xIndex] + 1) == 0);        // Copy remainder of longer number        while (xIndex > 0)            result[--xIndex] = x[xIndex];        // Grow result if necessary        if (carry) {            int bigger[] = new int[result.length + 1];            System.arraycopy(result, 0, bigger, 1, result.length);            bigger[0] = 0x01;            return bigger;        }        return result;    }

3、如果是减法

    /**     * Subtracts the contents of the second int arrays (little) from the     * first (big).  The first int array (big) must represent a larger number     * than the second.  This method allocates the space necessary to hold the     * answer.     */    private static int[] subtract(int[] big, int[] little) {        int bigIndex = big.length;        int result[] = new int[bigIndex];        int littleIndex = little.length;        long difference = 0;        // Subtract common parts of both numbers        while(littleIndex > 0) {            difference = (big[--bigIndex] & LONG_MASK) -                         (little[--littleIndex] & LONG_MASK) +                         (difference >> 32);            result[bigIndex] = (int)difference;        }        // Subtract remainder of longer number while borrow propagates        boolean borrow = (difference >> 32 != 0);        while (bigIndex > 0 && borrow)            borrow = ((result[--bigIndex] = big[bigIndex] - 1) == -1);        // Copy remainder of longer number        while (bigIndex > 0)            result[--bigIndex] = big[bigIndex];        return result;    }

4、输出就默认转成十进制

小结一下：

我花的时间大概在一个小时，代码行数在70行上下，没反转，尽量不存数据，中间处理过程细节多了一点，面试的时候可能要注意的是：

0、漏了一种边界情况，输入的大数是负数

1、想清楚再动手写代码

2、不要太扣细节，保存简单快速

最近找了一些leetcode的题在看，发现有些题目是需要思考比较久的，还是要刷算法题，至少知道解决的可能性思路。

上周面试官还问了一个内存泄漏的问题，当时没get到他要问的是什么，我就回答了一下我所经历过的内存泄漏。刚才突然想到，他问的问题可能是，有哪些情况会导致在Activity中导致内存泄漏。

1、如果在生命周期比较长的，例如service中保持activity的引用

2、或者是在static中保持activity的引用

3、或者是在内部类、线程中保存activity的引用

都有可能导致activity的内存泄漏

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