摘要：//Delete a specific char from string,,,,char is 'F' in string= "YOUSUF".the resultant string will be "YOUSU".#include #include #include using namespace std;void DeleteCharFromString(string* str, char c) { if (!str || str->empty()) { throw exception(); } int pos = 0; in 阅读全文

摘要：//example:Str="4142434546" Findout missing no 44.Add it to str;//Output:"414243444546". #include #include bool FixStr(const string& str, string* result) { if (str.empty()) return false; int len = str.length(); int first_len = 1; for (; first_len len) { matched = false; break; 阅读全文

摘要：dp，时间O（m*n），m为目标值，n为硬币数，空间O（m）/* * Given a list of 'N' coins, their values being in an array A[], return the minimum number of coins required to sum to 'S' (you can use as many coins you want). If it's not possible to sum to 'S', return -1 * * * Input #01: * Coin denomina 阅读全文

摘要：//Write a function to check if two rectangles defined as below, have common area or not. The functions take the top left and bottom right coordinate as input and return 1 if they have common area, otherwise return 0.struct Point { int x; int y;};struct Rec { Point l_top; Point r_bot;};bool Che... 阅读全文

摘要：O(n)的算法#include #include #include #include using namespace std;int LongestPlalindrome(const string& str) { if (str.empty()) return 0; string recombined_str; max_len = 1; recombined_str.resize(2 * str.size() - 1); for (int i = 0; i pld_len; pld_len.resize(recombined_str.size(), 1); int farres... 阅读全文

摘要：// 检查一颗树是否为二叉查找树#include #include using namespace std;#define MIN_NUM -0x80000000struct Node { int val; Node* left; Node* right;};bool CheckBinarySearchTree(Node* root) { if (!root) return false; stack s; int pre = MIN_NUM; Node* cur = root; while (cur || !s.empty()) { while (cur) { ... 阅读全文

摘要：擦，花了很久才调成功，push_back放到了循环外，真tm想抽自己，烦躁！先排序再做，效率会比较高，同时避免输出重复，对于重复数字，如果有n个重复数字，使用dp就是从包含0个到包含n个该数字的所有组合。// 输出数组中和为某个整数的所有组合#include #include #include using namespace std;void Print(const vector& arr) { for (int i = 0; i > Process(const vector& arr, int pos, int sum) { vector > results; i 阅读全文

摘要：很简单，思路一定要清晰。//数组有N-2个数字，数字的范围为1 ... N，没有重复的元素，要求打印缺少的2个数字, 空间复杂度O（1）。 #include using namespace std;void PrintNotExistNum(int arr[], int n) { bool end = false; bool end_pre = false; for (int i = 0; i < n; ++i) { if (arr[i] == -1 || arr[i] == i + 1) continue; while (arr[i] != -1 && arr[i] ! 阅读全文

摘要：利用数组存储，如果重复了在对应位置设置为-1,逻辑上考虑清楚，代码尽量清晰#include using namespace std;//数组有N+M个数字, 数字的范围为1 ... N, 打印重复的元素, 要求O(M + N), 空间复杂度O（1）。 void PrintRepeatedNum(int arr[], int n) { for (int pos = 0; pos < n; ++pos) { if (arr[pos] == -1) continue; while (arr[pos] != pos + 1) { int swap_pos = arr[pos] ... 阅读全文