需要注意overflow，特别是Integer.MIN_VALUE这个数字。
需要掌握二分法。
不用除法的除法，分而治之的乘方
2. Add Two Numbers
You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
基本等于合并数组，记得要最后加上carry位。
不需要考虑overflow，但是记得当输入两个空链表的时候输出0而不是空链表。
29. Divide Two Integers
Divide two integers without using multiplication, division and mod operator.
If it is overflow, return MAX_INT.
不能使用除法的除法，让除数不停地乘以二直到会大于被除数为止。
核心代码如下：
while (x >= y) {
int a = 0;
while (x >= (y << a)) {
a++;
}
a--;
result += (long)1 << a;
x -= y << a;
}
其中，x是被除数，y是除数，这段代码里，还需注意x,y都是long。1要被强制转换为long
这道题有很多corner case，具体见
[leetcode] 29. divide two integers
43. Multiply Strings
Given two numbers represented as strings, return multiplication of the numbers as a string.
Note:
- The numbers can be arbitrarily large and are non-negative.
- Converting the input string to integer is NOT allowed.
- You should NOT use internal library such as BigInteger.
- 按照我目前的解法没有什么corner case需要考虑，只要把输出数字前面多余的0去掉就可以了。但是现在的解法比较慢。有空的时候再改进吧。
50. Pow(x, n)
Implement pow(x, n).
1. 0的0次方是1
2. 判断x是否为0，n是否为0，x是否是1
3. 判断x的正负性，判断n的正负性。
4. 注意n是Integer.MIN_VALUE的情况，这种情况下取绝对值的时候会overflow
BTW，Double.MIN_VALUE定义的是大于0的最小正数哦~
leetcode的testcase没有涉及到double overflow的情况，double overflow的时候直接返回MAX_VALUE了？
69. Sqrt(x)
Implement int sqrt(int x).
Compute and return the square root of x.
用二分法做，因为测试 mid * mid 时得到的结果可能超过int的范围，所以，这一步需要用long来做。
int相乘overflow的时候，得到的结果有可能是一个正整数，无法判别overflow(得到的结果甚至可能大于mid真的无法判别啊)。
62. Unique Paths
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
How many possible unique paths are there?
Above is a 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
这道题其实是求组合数。Cm-1m+n-1
注意由于起点和终点一个在1一个在m，所以两者之间需要的步数是m - 1而不是m。
372. Super Pow
Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array.
Example1:
a = 2
b = [3] Result: 8
Example2:
a = 2
b = [1,0] Result: 1024
Credits:
Special thanks to @Stomach_ache for adding this problem and creating all test cases.
public class Solution {
private static final int N = 1337;
public int superPow(int a, int[] b) {
int size = b.length;
int result = 1;
for (int i = 0; i < size; i++) {
result = helper(result, 10) * helper (a, b[i]) % N;
}
return result;
}
private int helper(int a, int b) {
if (b == 0) {
return 1;
} else if (b == 1) {
return a % N;
}
return helper (a, b / 2) * helper(a, b - b / 2) % N;
}
}