LeetCode解题报告—— Linked List Cycle II & Reverse Words in a String & Fraction to Recurring Decimal

1. Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Note: Do not modify the linked list.

思路：想法是利用两指针，一个每次移动一步，另一个每次移动两步，如果存在环则这两个指针一定会相遇(这里可以在纸上画一下，因为后一个指针移动比前一个指针快，当后一个指针在环中来到前一个指针的前面时只有两种情况，在它前一个或者前两个，无论那种情况都能走到一起)。但是并不知道具体会在第几圈相遇，现在到了数学分析的时候了：https://leetcode.com/problems/linked-list-cycle-ii/discuss/44793/O(n)-solution-by-using-two-pointers-without-change-anything

    ListNode *detectCycle(ListNode *head) {
    if (head == NULL || head->next == NULL) return NULL;

    ListNode* firstp = head;
    ListNode* secondp = head;
    bool isCycle = false;

    while(firstp != NULL && secondp != NULL) {
        firstp = firstp->next;
        if (secondp->next == NULL) return NULL;
        secondp = secondp->next->next;
        if (firstp == secondp) { isCycle = true; break; }
    }

    if(!isCycle) return NULL;
    firstp = head;
    while( firstp != secondp) {
        firstp = firstp->next;
        secondp = secondp->next;
    }

    return firstp;
}

2. Reverse Words in a String

Given an input string, reverse the string word by word.

Example:

Input: "the sky is blue",
Output: "blue is sky the".

Note:

• A word is defined as a sequence of non-space characters.
• Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
• You need to reduce multiple spaces between two words to a single space in the reversed string.

String[] parts = s.trim().split("\\s+");
String out = "";
for (int i = parts.length - 1; i > 0; i--) {
    out += parts[i] + " ";
}
return out + parts[0];

注意的是上面的正则表达式的用法，\s 匹配任何不可见字符，包括空格、制表符、换页符等等。等价于[ \f\n\r\t\v]，后面的 + 表示匹配一个或多个，然后就是在java中使用正则前面要多加个反斜杠。

3. 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.

Example 1:

Input: numerator = 1, denominator = 2
Output: "0.5"

Example 2:

Input: numerator = 2, denominator = 1
Output: "2"

Example 3:

Input: numerator = 2, denominator = 3
Output: "0.(6)"

思路：难点在于小数部分的判断，可以用一个map，key是当前的被除数，value是当前计算结果的数字长度，如果在计算过程中key和之前的key相等，因为除数一直不变，所以这是开始循环，那么就要在上一个key所对应的value处插入 ( 符号，然后在末尾插入 ) 符号。

public class Solution {
    public String fractionToDecimal(int numerator, int denominator) {
        if (numerator == 0) {
            return "0";
        }
        StringBuilder res = new StringBuilder();
        // "+" or "-"
        res.append(((numerator > 0) ^ (denominator > 0)) ? "-" : ""); // 判断正负，异号的情形才是负
        long num = Math.abs((long)numerator);
        long den = Math.abs((long)denominator);

        // integral part
        res.append(num / den);　　// 先求出整数部分
        num %= den;　　// 求余数
        if (num == 0) {
            return res.toString();
        }

        // fractional part
        res.append(".");
        HashMap<Long, Integer> map = new HashMap<Long, Integer>();
        map.put(num, res.length());  // 将余数作为key加入到map中
        while (num != 0) {
            num *= 10;
            res.append(num / den);  // 上面num乘10是因为这里要取小数点最后一位的后一位的除后的整数，否则num/den就一直是0
            num %= den;　　// 求余数
            if (map.containsKey(num)) { // 如果余数和某一轮的被除数一样，因为除数一直不变，那么则表明开始无限循环了
                int index = map.get(num); // 取开始发生无限循环时res的长度已确定插入 ( 符号的位置
                res.insert(index, "(");
                res.append(")");
                break;
            }
            else {
                map.put(num, res.length()); // 如果没开始无限循环，则要记录当前除后结果数字的长度，以记录插入括号的位置
            }
        }
        return res.toString();
    }
}