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Arrays
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/*Program which takes an input array and an index and rearranges the array such that values less than the element come first, | |
elements equal follow and finally followed by elements greater than the index element*/ | |
//Notes: This is just the quicksort algorithm where they have given the random value and there is just one step | |
This solution is the same as the precvious code: only just add a condition to increment the current index when the values match | |
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/*To work with arrays in place we need to keep 2 pointers: first where the last evenm element was inserted and the next where | |
the last odd element was inserted. We can use a vector for the array here.*/ | |
#include <vector> | |
void swap(int &a,int &b) | |
{ | |
int temp = a; | |
a = b; | |
b = temp; | |
} | |
void sortEvenOdd(vector<int>* array) | |
{ | |
int evenIndex = 0,currentIndex = 0; | |
int oddIndex = array.size()-1; | |
while (currentIndex < oddIndex) | |
{ | |
if (array[currenIndex] % 2 == 0) | |
{ | |
swap(array[currentIndex],array[evenIndex]); | |
evenIndex++; | |
} | |
else | |
{ | |
swap(array[currentIndex],array[oddIndex]); | |
oddIndex++; | |
} | |
currentIndex++; | |
} | |
} | |
//Notes: CurrentIndex is not really needed. You can just use the even index! |
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/*Program to take an array representing a decimal and adding 1 to it | |
eg. <1,2,9> will give result <1,3,0> | |
*/ | |
vector<int> AddOne(vector<int> A) | |
{ | |
++A.back(); | |
for (int i = A.size() -1; A[i] == 10 && i > 1; i++) | |
{ | |
A[i] = 0; | |
++A [i-1]; | |
} | |
if (A[0] == 10) | |
{ | |
A[0] = 0; | |
A.insert(A.begin(),1); | |
} | |
return A; | |
} |
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