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목록Leetcode (18)
Swimmer
Solution Use Queue First time to use C++ STL, it is very useful than coding data structure one by one in C #include #include // BFS Solution // Use Queue // Use STL -> Easy typedef struct myStruct { int RowIdx; int ColumnIdx; int depth; }myStruct; using namespace std; void BFS(int row, int column, queue* pmy_queue, vector* pmyvector); int gArr[100][100] = { 0 }; int main() { queue my_queue; vect..
Solution Use ASCII transformation lower case english letter to integer #include int GetNumericValue(char* ArrWord); bool isSumEqual(char* firstWord, char* secondWord, char* targetWord) { bool RetVal = true; int fisrtWordNumericValue = GetNumericValue(firstWord); int secondWordNumericValue = GetNumericValue(secondWord); int targetWordNumericValue = GetNumericValue(targetWord); if (targetWordNumer..
Solution - Use qsort(2D array, 2 column) & Greedy Algorithms - Time Limit Exceed int minGroups(int** intervals, int intervalsSize, int* intervalsColsize) { int i = 0; int** RefArr; int RefArrSize = intervalsSize; int** TmpArr; int TmpArrSize = intervalsSize; int NumOfGroup = 0; int StartIndex = 0; int EndIndex = 0; // Sort qsort(intervals, intervalsSize, sizeof(intervals[0]), compare2Darray2Colu..
Solution - Simple Count of Max Depth Time Complexity O(n) Space Complexity O(n) int maxDpeth(char* s); int main() { char string[] = "(1)+((2))+(((3)))"; int RetVal; RetVal = maxDpeth(string); return 0; } int maxDpeth(char* s) { int Depth = 0; int MaxDepth = 0; int NumOfChar = 0; while (s[NumOfChar] != '\0') { if (s[NumOfChar] == '(') { ++Depth; } else if (s[NumOfChar] == ')') { --Depth; } else {..