may you please help me in solving any one of these three questions? the three are related to Data Structures and Algorithms with object-oriented design patterns in java
Q1)Devise a scheme using a stack to convert an infix expression to a postfix expression.Hint: in a postfix expression operators appear after their operands, whereas in an infix expression they appear between their operands.Process the symbols in the prefix one by one.Output operands immediately, but save the operators in a stack until they are needed.Pay special attention to the precedence of the poerators.The program must catch appropriate exceptions and display informative error messages when the input is in error.
Q2. Write the following recursive methods to be as general as possible. (a) Write a recursive method public int countGreaterThan(int n) that returns the number of integers in a given one-D array that are greater than the integer n. Use this method to write another recursive method public int countGreaterThan2D(int n) that returns the number integers in a two-dimensional array that are greater than the integer n. (b) Write a complete program to test your method using the following array: int twoD[][]={{-11,-10,-9},{-8,-7,-6,-5,-4,-3,-2},{-1,0},{1,2,3,4,5,6,7}, {8,9,10,1 1},{}}; Your program should behave as exemplified below: Please enter an integer: -1 There are 12 numbers greater than -1 in the array.
Q3. Study the following methods carefully. Determine the big-O running time of the method, myMethod() by forming and solving recurrence relations. class Hw3Q3{ int helperArray[]; // ... assume space has been allocated for this array as needed public static void myMethod(int array[], int i, int n){ if(n>1){ myMethod(array, i, n/2); myMethod(array, i + n/2, n-n/2); helperMethod(array, i, n/2, n-n/2); } public static void helperMethod(int []a, int left, int mid, int right){ int i = left; int j = left; int k = mid + 1; while(j<=mid && k<=right){ if(array[j]<array[k]) helperArray[i++]=array[j++]; else helper Array[i++]=array[k++]; } while(j<=mid) helperArray[i++]=array[j++]; for(i=lef t; i<k; ++i) array=helperArray; } } }
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