sinat_29844779的博客

struct BinaryTreeNode { int val; BinaryTreeNode* left; BinaryTreeNode* right; }; 一、判断二叉树B是不是A的子树 bool DoseTree1HasTree2(BinaryTreeNode* pRoot1,BinaryTreeNode* pRoot2) { if(pRoot2==NULL) return

方法一、p[j+1]=='*'，此时从s[i]开始的子串，假设是s[i],s[i+1].....s[i+k]都等于p[j]，则意味着这些都可能是合适的匹配，这就得递归剩下的（i,j+2）,(i+1,j+2).......(i+k,j+2) 实现代码：bool isMatch(string s, string p) { return check(s,p,0,0); }

'?' Matches any single character. '*' Matches any sequence of characters (including the empty sequence). The matching should cover the entire input string (not partial). The function prototype shoul

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possible, it must rearrange it as the lowest possible

今天学习了一个非常高效的数据结构，那就是跳表。其查找效率比红黑树要更高。整个的实现见下面：//参考：Skip lists: a probabilistic alternative to balanced trees #ifndef _SKIPLIST_H_ #define _SKIPLIST_H_ #include #include #include #include #define

Write a program to solve a Sudoku puzzle by filling the empty cells. Empty cells are indicated by the character '.'. 这道题是一个回溯法来解决的问题，依次试探每一种可能的情况，当满足这一情况时再向后递归直至整个问题解决。真正理解了回溯法的强大 啊，感谢啊 。 void

The set [1,2,3,…,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order, We get the following sequence (ie, for n = 3): "123""132""213""231""3

Implement int sqrt(int x). Compute and return the square root of x. 好精妙的方法啊 int mySqrt(int x) { long r = x; while (r*r > x) r = (r + x/r) / 2; return r;

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). For example, S = "ADOBECODEBANC" T = "ABC" Minimum window is "BAN

这种类型的题目有两种， 第一种情况下可以对里面的数字重复使用， vector> vec; vector> combinationSum(vector& candidates, int target) { sort(candidates.begin(),candidates.end()); vector local; check(candi

//最长子序列 int longestPalindromeSubseq(string s) { int size = s.size(); vector> dp(size,vector(size,0)); for(int i=0;i<size;i++) dp[i][i]=1; for(int len=1;len<

struct TreeNode { int val; TreeNode *left; TreeNode *right; }; void PreOrder(TreeNode *root) { if(root==NULL) return; stack stk; TreeNode *p = root; while(p!=NULL||stk.empty()) { while(p!=

void PrintMatrixIncircle(int** numbers,int columns,int rows,int start) { int endX = columns-1-start; int endY = rows-1-start; for(int i = start;i<=endX;i++) { int number = numbers[start][i];

void PrintNum(char* number) { bool Beginning0=true; for(unsigned int i = 0;i<strlen(number);i++) { if(Beginning0&&number[i]!='0') Beginning0 = false; if(!Beginning0) printf("%c",number[i]

//一个数组里面的所有的数字出现两次，只有一个数只出现一次，找出这个数 void FindUniqueNum(vector p,int num) { num = 0; for(int i = 0;i<p.size();i++) num^=p[i]; } //一个数组里面的其余的数出现两次，有两个不同的数出现一次，找出这两个数 void FindTwoUniqueNum(vector p,i

struct ListNode { int val; ListNode *next; ListNode(int x):val(x),next(NULL){} }; //从一个链表中删除重复的节点，保留一个该节点 ListNode* deleteDuplicates(ListNode* head) { if(head==NULL) return head; ListNode *pre

ListNode* Merge(ListNode* p1,ListNode* p2) { if(p1==NULL) return p2; if(p2==NULL) return p1; if(p1->valval) { p1->next = Merge(p1->next,p2); return p1; } else { p2->next

void QSort(int *num,int low,int high) { int pivot; if(low<high) { pivot=Partition(num,low,high); QSort(num,low,pivot-1); QSort(num,pivot+1,high); } } int Partition(int *num,int low,int

//硬币的找零问题，给定一个数值，求使得最少的硬币数组合成所要的数值 int CoinChange(vector coins,int len,int sum) { vector dp(sum+1,sum+1); dp[0]=0;//设定为0 for(int i=1;i<=sum;i++) { for(int j=0;j<len;j++) { if(coins[j]<=i)

'.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). The function prototype should be: bool isMatch(c