Sort Integers

Sort Integers

Given an integer array, sort it in ascending order. Use selection sort, bubble sort, insertion sort or any O(n²) algorithm.

Example

Given [3, 2, 1, 4, 5], return [1, 2, 3, 4, 5].

selection sort

From A[0] to A[A.length - 1], find the minimum and swap with A[0].
From A[1] to A[A.length - 1], find the minimum and swap with A[1].
From A[2] to A[A.length - 1], find the minimum and swap with A[2].
...

public class Solution {
    /**
     * @param A an integer array
     * @return void
     */
    public void sortIntegers(int[] A) {
        // Write your code here
        for (int i = 0; i < A.length - 1; i++) {
            int min = A[i];
            int minIndex = i;
            for (int j = i + 1; j < A.length; j++) {
                if (A[j] < min) {
                    min = A[j];
                    index = j;
                }
            }
            int tmp = A[i];
            A[i] = A[minIndex];
            A[minIndex] = tmp;
        }
    }
}

insertion sort

From A[1], if A[1] < A[0], swap them.
From A[2], if A[2] < A[1], swap them. Now if A[1] < A[0], swap them.
From A[3], if A[3] < A[2], swap them. Now if A[2] < A[1], swap them. Now if A[1] < A[0], swap them.
...
Every time A[i] compares with its previous element and inserts to the proper position.

public class Solution {
    /**
     * @param A an integer array
     * @return void
     */
    public void sortIntegers(int[] A) {
        // Write your code here
        for (int i = 1; i < A.length; i++) {
            int currIndex = i;
            while (currIndex != 0 && A[currIndex] < A[currIndex - 1]) {
                int tmp = A[currIndex];
                A[currIndex] = A[currIndex - 1];
                A[currIndex - 1] = tmp;
                currIndex--;
            }
        }
    }
}

bubble sort

From A[1] to A[A.length - 1], if A[i - 1] > A[i], swap them. Now A[A.length - 1] is the maximum number.
From A[1] to A[A.length - 2], if A[i - 1] > A[i], swap them. Now A[A.length - 2] is the 2nd maximum number.
From A[1] to A[A.length - 3], if A[i - 1] > A[i], swap them. Now A[A.length - 3] is the 3rd maximum number.

...

public class Solution {
    /**
     * @param A an integer array
     * @return void
     */
    public void sortIntegers(int[] A) {
        // Write your code here
        for (int i = 0; i < A.length; i++) {
            for (int j = 1; j < A.length - i; j++) {
                if (A[j - 1] > A[j]) {
                    int tmp = A[j - 1];
                    A[j - 1] = A[j];
                    A[j] = tmp;
                }
            }
        }
    }
}