xjwowangran的专栏

题目如下： Sort a linked list in O(n log n) time using constant space complexity. nlogn的排序有快速排序、归并排序、堆排序。双向链表用快排比较适合，堆排序也可以用于链表，单向链表适合用归并排序。以下是用归并排序的代码：  #include "stdafx.h" #include using nam

Subset class Solution { public: vector > subsets(vector &S) { vector > result; if(S.size() == 0) return result; sort(S.begin(), S.end()); int n = S.size(); vector each; resul

Add Two Numbers 主要考虑两个链表相等位数的和，shengxia /* 提交多次才成功，主要是考虑，两个链表不一样长度 1. 12 7990，有数有进位 2. 1 99，无数有进位 5 5， */ class Solution { public: ListNode *addTwoNumbers(ListNode *l1, ListNode *l2

3 Sum vector > threeSum(vector &num) { vector > result; if(num.size() < 3) return result; sort(num.begin(),num.end()); const int target = 0; auto last = num.end(); for(auto a = num.begin();

Pascal's Triangle   class Solution { public: vector > generate(int numRows) { vector > result; if(numRows <= 0) return result; vector row1(1,1); vector row2(2,1); if(numRows >= 1) re

class Solution { public: vector plusOne(vector &digits) { if(digits.size() == 0) return digits; int carry = 0; for(int i = digits.size()-1; i >=0 ; -- i) //如果使用的是size_t就是无符号的，如果减一了，那么就是一个大数

Minimum Path Sum class Solution { public: vector > f; //备忘录法，把每次计算的值保存下来 int getOrupdate(vector > &grid, int m, int n) { if(m < 0 || n < 0) retu

Validate Binary Search Tree   浪费太多时间思考这个问题，结果知道了左孩子的右孩子必须比爷爷小，右孩子的左孩子必须比爷爷大，可是这么递归是超时的，实际上是归纳成先序遍历，递归判断，递归时传入两个参数，一个是左界，一个是右界，节点的值必须在两个界的中间，同时在判断做子树和右子树时更新左右界。 class Solution { public: bool Va

Container With Most Water   Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of li

Search a 2D Matrix   Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted from left to ri

Binary Tree Level Order Traversal class Solution { public: vector > levelOrder(TreeNode *root) { vector > result; vector temp; queue myQueue; TreeNode* p

主要是卡特兰特s class Solution { public: int numTrees(int n) { if(n == 0) return 0; int* num = new int[n+1]; num[0] = 1; num[1] = 1; for(

bool check(TreeNode* left, TreeNode* right) { if(left == NULL && right == NULL) return true; if(left == NULL ||right == NULL) return false; if(left->val == right->val) return check(left->left

1.same tree /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };

class Solution { public: void nextPermutation(vector &num) { if(num.empty()) return; size_t j = num.size() - 1; size_t i = j - 1; while(i >= 0 && j > 0 && num[j] <= num[i]) { -- j; -- i;

if(head == NULL || head->next == NULL) return head; else { ListNode* cur = head->next; ListNode* cprior = head; for(;cur != NULL; cprior = cur, cur = cur->next) { ListNode* index = he

/** * Definition for binary tree with next pointer. * struct TreeLinkNode { * int val; * TreeLinkNode *left, *right, *next; * TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL)

class Solution { public: bool isValid(vector& board, int n,int x, int y) { int i =0, j =0; for(i=0; i < x; ++ i) { if(board[i][y] == 'Q')//只可能是x之前的行，在同一列有相同的元素 return false; } for(i = 0; i

Remove Duplicates from Sorted Array   class Solution { public: int removeDuplicates(int A[], int n) { if(n<=0) return 0; if(n==1) return 1; int i=0,j=1; //遍历一边又相同的就用后面的元素覆盖 whil

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: bool hasCycl

Convert Sorted Array to Binary Search Tree

Longest Consecutive Sequence

// KMP.cpp : 定义控制台应用程序的入口点。 // #include "stdafx.h" #include #include using namespace std; //输入：src, slen主串 //输入：pstr, plen模式串 //输入：next KMP算法中的next函数值数组 //修正后的求next数组各值的函数代码 void GetNext(

class Solution { public: bool isPalindrome(string s) { if(s.empty() ) return true; size_t i = 0; size_t j = 0; string s1 = ""; for(; i < s.size(); ++ i) { if('A'= s[i] ) { s1 +=

Search in Rotated Sorted Array

wewr

Generate Parentheses   Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = 3, a solution set is: "((()))", "(()