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();
}
}
