Algorithm: 计算类题目归类整理

提纲：
166. Fraction to Recurring Decimal

43. Multiply Strings
415. Add Strings

67. Add Binary（思路和模板与上一题相同）
371. Sum of Two Integers（用位操作来代替实际的加减）
下面两题思路一样（找到第一个非9的数，接着执行相应的操作）：
66. Plus One
369. Plus One Linked List
注：链表加法的两道题目归为链表类

166. Fraction to Recurring Decimal
Given two integers representing the numerator and denominator of a fraction, return the fraction in string format.

If the fractional part is repeating, enclose the repeating part in parentheses.

For example,

Given numerator = 1, denominator = 2, return "0.5".
Given numerator = 2, denominator = 1, return "2".

Given numerator = 2, denominator = 3, return "0.(6)".
思路：用一个hash table去记录从小数点开始的余数。因为两者都是整数，所以小数点之后都是0. 因为，如果从某个点开始重新遇到某个之前见过的余数（remainder）代表循环小数的部分出现了。
Extreme cases: 分子numerator是0， 分母dominator是0. 负数是Integer.MIN_VALUE的时候会溢出。
功能测试：两者其中之一或者两者是负数。两者是整数，结果是整数、结果是小数（有限或者无限）。public String fractionToDecimal(int numerator, int denominator) {
if (numerator == 0) {
return "0";
}
StringBuilder fraction = new StringBuilder();
// If either one is negative (not both)
if (numerator < 0 ^ denominator < 0) {
fraction.append("-");
}
// Convert to Long or else abs(-2147483648) overflows
long dividend = Math.abs(Long.valueOf(numerator));
long divisor = Math.abs(Long.valueOf(denominator));
fraction.append(String.valueOf(dividend / divisor));
long remainder = dividend % divisor;
if (remainder == 0) {
return fraction.toString();
}
fraction.append(".");
Map<Long, Integer> map = new HashMap<>();
while (remainder != 0) {
if (map.containsKey(remainder)) {
fraction.insert(map.get(remainder), "(");
fraction.append(")");
break;
}
map.put(remainder, fraction.length());
remainder *= 10;
fraction.append(String.valueOf(remainder / divisor));
remainder %= divisor;
}
return fraction.toString();
}
43. Multiply Strings

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2.
Note:
The length of both num1 and num2 is < 110.
Both num1 and num2 contains only digits 0-9.
Both num1 and num2 does not contain any leading zero.
You must not use any built-in BigInteger library or convert the inputs to integer directly.
思路：两个数相乘得到的数字，比最多不会超过两个数的长度加起来的长度。例如（99 * 99 < 10000）
所以，StringBuilder的长度是这个长度。最后删除首部的0。接下来是正常的乘法，乘数每一位分别和被乘数的每一位相乘，用嵌套的for-loop即可实现。注意结果的坐标是i + j + 1，考虑到9 * 9或者1 * 1即可知。

public class Solution {
public String multiply(String num1, String num2) {
int n1 = num1.length(), n2 = num2.length();
int[] products = new int[n1 + n2];
for (int i = n1 - 1; i >= 0; i--) {
for (int j = n2 - 1; j >= 0; j--) {
int d1 = num1.charAt(i) - '0';
int d2 = num2.charAt(j) - '0';
products[i + j + 1] += d1 * d2;
}
}
int carry = 0;
for (int i = products.length - 1; i >= 0; i--) {
int tmp = (products[i] + carry) % 10;
carry = (products[i] + carry) / 10;
products[i] = tmp;
}
StringBuilder sb = new StringBuilder();
for (int num : products) sb.append(num);
while (sb.length() != 0 && sb.charAt(0) == '0') sb.deleteCharAt(0);
return sb.length() == 0 ? "0" : sb.toString();
}
}

415. Add StringsGiven two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2.
Note:
The length of both num1 and num2 is < 5100.
Both num1 and num2 contains only digits 0-9.
Both num1 and num2 does not contain any leading zero.
You must not use any built-in BigInteger library or convert the inputs to integer directly.

思路：最长只会比两个数之间长度较长的那个数多一位。接下来就是普遍的加法需要注意的问题，carry必须最后检查一次并加上和while循环的时候记得递增或者递减变量。与链表加法的模板类似。

class Solution {
public String addStrings(String num1, String num2) {
int l1 = num1.length();
int l2 = num2.length();

int carry = 0;
int i = l1 - 1;
int j = l2 - 1;
int res = 0;

StringBuilder sb = new StringBuilder();
while (i >= 0 && j >= 0) {
res = (num1.charAt(i--) - '0') + (num2.charAt(j--) - '0') + carry;
carry = res / 10;
res = res % 10;
sb.append(Integer.toString(res));
}

while (i >= 0) {
res = (num1.charAt(i--) - '0') + carry;
carry = res / 10;
res = res % 10;
sb.append(Integer.toString(res));
}

while (j >= 0) {
res = (num2.charAt(j--) - '0') + carry;
carry = res / 10;
res = res % 10;
sb.append(Integer.toString(res));
}

if (carry > 0) {
sb.append(Integer.toString(carry));
carry = 0;
}

return sb.reverse().toString();
}
}

67. Add Binary
Given two binary strings, return their sum (also a binary string).

For example,
a = "11"
b = "1"
Return "100".

public class Solution {
public String addBinary(String a, String b) {
StringBuilder sb = new StringBuilder();
int i = a.length() - 1, j = b.length() -1, carry = 0;
while (i >= 0 || j >= 0) {
int sum = carry;
if (j >= 0) sum += b.charAt(j--) - '0';
if (i >= 0) sum += a.charAt(i--) - '0';
sb.append(sum % 2);
carry = sum / 2;
}
if (carry != 0) sb.append(carry);
return sb.reverse().toString();
}
}

371. Sum of Two IntegersCalculate the sum of two integers a and b, but you are not allowed to use the operator + and -.
Example:
Given a = 1 and b = 2, return 3.

/**
Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.

Example:
Given a = 1 and b = 2, return 3.

a作为直接加的数，b作为进位，继续递归。
*/
class Solution {
public int getSum(int a, int b) {
if (b == 0) {
return a;
}
int newA = a ^ b;
int newB = (a & b) << 1;
return getSum(newA, newB);
}
}
369. Plus One Linked List
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode plusOne(ListNode head) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode lastNotNine = dummy, node = head;

while (node != null) {
if (node.val != 9) {
lastNotNine = node;
}
node = node.next;
}
lastNotNine.val++;
node = lastNotNine.next;
while (node != null) {
node.val = 0;
node = node.next;
}
return dummy.val == 1 ? dummy : dummy.next;
}
}
66. Plus Oneclass Solution {
public int[] plusOne(int[] digits) {
if (digits == null || digits.length == 0) {
return digits;
}

int lastNotNine = -1;
for (int i = digits.length - 1; i >= 0; --i) {
if (digits[i] != 9) {
lastNotNine = i;
break;
}
}

if (lastNotNine == -1) {
int[] ans = new int[digits.length + 1];
ans[0] = 1;
return ans;
}

int zeroIndex = lastNotNine + 1;
while (zeroIndex != digits.length) {
digits[zeroIndex++] = 0;
}

digits[lastNotNine]++;
return digits;
}
}