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Merge pull request #3 from algorithm001/master · algorithm001/algorithm@20515e2 · GitHub

/ algorithm Public

Commit 20515e2

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fishLite
authored
Merge pull request #3 from algorithm001/master
update forks
2 parents 03bc0ab + 455c09c commit 20515e2

19 files changed

Lines changed: 617 additions & 2 deletions

‎.gitignore‎

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.idea
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.vscode

‎Week_01/id_0/TestPush.java‎

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public class TestPush {
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public static void main(String[] args) {
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System.out.println("Hello Geekbang~~");
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}
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}

‎Week_01/id_102/leetcode_153_102.cpp‎

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/*
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* 存两个临时变量，left和right, 如果left>right,则返回right;
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* 如果只有一个元素，则返回它自己；
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* 如果没有旋转过，则返回第一个元素
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*/
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class Solution {
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public:
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int findMin(vector<int>& nums) {
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int count = nums.size();
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if (count == 1) {
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return nums[0];
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}
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int left = nums[0];
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int right = nums[1];
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if (left > right) {
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return right;
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}
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int i;
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for (i = 2; i < count; i++) {
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left = right;
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right = nums[i];
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if (left > right) {
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return right;
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}
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}
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if (i == count) {
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return nums[0];
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}
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return -1;
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}
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};
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‎Week_01/id_102/leetcode_20_102.cpp‎

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/*
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* 堆栈实现
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* 满足三个条件则有效
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* 1. “左” 压栈
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* 2. “右” 在栈顶查找匹配的括号, 如果找到弹出栈顶字符，否则返回false
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* 3. 最终"栈" 为空, 返回true, "栈" 非空, 返回false
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* 注意，对于第一次输入是')','}'或者']'的情况，一定要判断堆栈是否为空，否则会数组越界
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*/
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class Solution {
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public:
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bool isValid(string s) {
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vector<char> stack;
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map<char, char> spouse;
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spouse[')'] = '(';
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spouse['}'] = '{';
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spouse[']'] = '[';
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for (int i = 0; i< s.size(); ++i) {
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if (s[i] =='(' || s[i] =='[' || s[i] =='{') {
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stack.push_back(s[i]);
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} else {
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if (stack.empty() || spouse[s[i]] != stack[stack.size()-1]) {
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return false;
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}
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stack.pop_back();
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}
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}
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return stack.empty();
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}
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};
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‎Week_01/id_127/21.jpeg‎

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‎Week_01/id_127/83.jpeg‎

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‎Week_01/id_127/LeetCode_21_127.java‎

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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
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if (l1 == null)
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return l2;
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if (l2 == null)
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return l1;
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ListNode cur1 = l1;
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ListNode cur2 = l2;
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ListNode mergeCur = null;
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ListNode mergeHead = null;
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if (cur1.val <= cur2.val) {
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mergeCur = cur1;
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mergeHead = cur1;
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cur1 = cur1.next;
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} else {
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mergeCur = cur2;
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mergeHead = cur2;
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cur2 = cur2.next;
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}
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while (cur1 != null && cur2 != null) {
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if (cur1.val <= cur2.val) {
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mergeCur.next = cur1;
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cur1 = cur1.next;
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} else {
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mergeCur.next = cur2;
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cur2 = cur2.next;
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}
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mergeCur=mergeCur.next;
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}
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if (cur1 != null) {
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mergeCur.next = cur1;
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}
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if (cur2 != null) {
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mergeCur.next = cur2;
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}
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return mergeHead;
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}
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/**
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* 递归思路没有想到，参考网上的相关答案，理解之后写出
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*/
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public ListNode mergeTowListsRecursive(ListNode l1, ListNode l2) {
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if (l1 == null)
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return l2;
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if (l2 == null)
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return l1;
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if (l1.val <= l2.val) {
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ListNode head = l1;
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head.next = mergeTowListsRecursive(l1.next, l2);
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return head;
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} else {
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ListNode head = l2;
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head.next = mergeTowListsRecursive(l1, l2.next);
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return head;
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}
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}
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}

‎Week_01/id_127/LeetCode_83_127.java‎

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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public ListNode deleteDuplicates(ListNode head) {
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ListNode cur = head;
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while (cur != null && cur.next != null) {
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if (cur.val == cur.next.val) {
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cur.next = cur.next.next;
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} else {
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cur = cur.next;
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}
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}
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return head;
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}
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}

‎Week_01/id_127/NOTE.md‎

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# 学习笔记
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# 学习笔记
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### LeetCode-83思路：
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见图片: `83.jpeg`
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### LeetCode-21思路：
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‎Week_01/id_27/LeetCode_142_027.java‎

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/**
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* Definition for singly-linked list.
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* class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) {
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* val = x;
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* next = null;
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* }
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* }
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*/
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public class Solution {
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//暴力解法
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public ListNode detectCycle(ListNode head) {
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Set<ListNode> set = new HashSet<ListNode>();
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while (head != null) {
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if (set.contains(head))
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break;
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set.add(head);
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head = head.next;
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}
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return head;
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}
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public ListNode detectCycle2( ListNode head ) {
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if( head == null || head.next == null ){
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return null;
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}
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//1、寻找相遇点
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ListNode fp = head, sp = head;
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while( fp != null && fp.next != null){
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sp = sp.next;
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fp = fp.next.next;
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if( fp == sp ){
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break;
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}
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}
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//2、没有环的情况直接退出
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if( fp == null || fp.next == null ){
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return null;
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}
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//3、有环时，slow起点设置为head，fast从相遇点出发，相遇点即为入环点
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sp = head;
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while( fp != sp ){
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sp = sp.next;
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fp = fp.next;
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}
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return sp;
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}
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}

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