daisyZH的专栏

转自：http://www.cnblogs.com/dolphin0520/archive/2011/08/24/2151846.html在介绍KMP算法之前，先介绍一下BF算法。一.BF算法BF算法是普通的模式匹配算法，BF算法的思想就是将目标串S的第一个字符与模式串P的第一个字符进行匹配，若相等，则继续比较S的第二个字符和P的第二个字符；若不相等，则比较S的第二个字符和P的第一

class Solution {public:    int bsearch(int A[], int beg, int end, int target) {        if (beg>end) {            return -1;        }        int mid = (beg+end)/2;        if (A[mid]==targ

class Solution {public:    bool searchMatrix(vector > &matrix, int target) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int m = matrix

转载请注明原文网址： http://www.cnblogs.com/xianyunhe/archive/2011/11/26/2264709.html函数指针是通过指向函数的指针间接调用函数。函数指针可以实现对参数类型、参数顺序、返回值都相同的函数进行封装，是多态的一种实现方式。由于类的非静态成员函数中有一个隐形的this指针，因此，类的成员函数的指针和一般函数的指针的表现形式不一样。

转自：http://blog.youkuaiyun.com/martin_liang/article/details/8982562Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.For example,Given the

class Solution {public:    int core(int A[], int beg, int end, int target) {        if (beg==end && A[beg]            return beg+1;        }        if (beg==end&&A[beg]>target) {       

class Solution {public:    bool isValid(string s) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int n = s.size();        if (n==0) {

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

泪奔了，改了一个边界条件就华丽丽的通过了~在这之前，做了好多无用功，只想说一句，叉~class Solution {public:   string longestPalindrome(string s) {        // Start typing your C/C++ solution below        // DO NOT write int main() fu

class Solution {private:    bool canUse[256];public:   int lengthOfLongestSubstring(string s) {        // Start typing your C/C++ solution below        // DO NOT write int main() function

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

class Solution {public:    int removeDuplicates(int A[], int n) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (A==NULL || n       

class Solution {public:    int removeElement(int A[], int n, int elem) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (A==NULL||n 

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

class Solution {public:    int removeDuplicates(int A[], int n) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if(A==NULL||n         

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

转自：http://www.cnblogs.com/graph/archive/2013/07/31/3229056.htmlYou are given an n x n 2D matrix representing an image.Rotate the image by 90 degrees (clockwise).Follow up:

转自：http://blog.youkuaiyun.com/wyc1230/article/details/7263055Given a roman numeral, convert it to an integer.Input is guaranteed to be within the range from 1 to 3999.[cpp] view plaincop

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

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

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

转自：http://blog.unieagle.net/2012/09/29/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Ainterleaving-string%EF%BC%8C%E4%BA%8C%E7%BB%B4%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/LeetCode题目：Interleaving String，二维

转载：http://blog.youkuaiyun.com/zxzxy1988/article/details/8587244今天发现了leetcode上面一道题，觉得非常经典，记录之。题目是这样的：给定两个已经排序好的数组（可能为空），找到两者所有元素中第k大的元素。另外一种更加具体的形式是，找到所有元素的中位数。本篇文章我们只讨论更加一般性的问题：如何找到两个数组中第k大的元素？不过，

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

#include #include using namespace std;void dfs(int depth, vector& s, vector >& res, vector a, int start) {   res.push_back(a);   if (depth == s.size())       return;   for (int i = sta

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

class Solution {public:    bool equal(double x, double y) {        if (abs(x-y)             return true;        }        return false;    }    double pow(double x, int n) {        //

class Solution {public:    bool isPal(string s) {        if (s.size()            return true;        }        int beg=0;        int end=s.size()-1;        while (beg            if(s[

#include #include struct ListNode {     int val;     ListNode *next;     ListNode(int x) : val(x), next(NULL) {} };    ListNode *reverseBetween(ListNode *head, int m, int n) {       

class Solution {public:    int atoi(const char *str) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (str == NULL) {            retu

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

牛顿迭代法class Solution {public:    int sqrt(int x) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (x            return x;       

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

class Solution {public:    vector plusOne(vector &digits) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (digits.empty()) {        

class Solution {public:     vector getRow(int rowIndex) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector res;         res.push_

class Solution {public:    vector > generate(int numRows) {        // Start typing your C/C++ solution below        // DO NOT write int main() function             vector > res;        i

class Solution {public:    vector twoSum(vector &numbers, int target) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector res;     

class Solution {public:    vector grayCode(int n) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int size = 1        vector res;   

class Solution {public:    int minDistance(string word1, string word2) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int len1 = word1.s

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