Sorting numbers in java

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melmantheman	Created : 02 November 2011 Edited : 02 November 2011 System : Cross Platform Language : Ruby Sorting numbers in java This can be used to sort high scores. it goes least to greatest but a simple turn around isn't to hard.
//Erik Decker import java.util.; import java.io.; public class Sorts { /** * Constructor for objects of class Sorts / public Sorts() { } /* * sorts the array using bubble sort method * @param nums the array of numbers * @return int[] the sorted array / public int[] bubbleSort(int[] nums) { for(int x = 0;x<nums.length;x++) { for(int x2 = 1; x2 < nums.length -x;x2++) { if(nums[x2-1] > nums[x2]) // later one is greater than earlier one { int t = nums[x2-1]; // store the one at x2 as temp nums[x2-1] = nums[x2]; // replace with x nums[x2] = t; //make x's into x2's } } } return nums; } /* * sorts the array using selection sort method * @param nums the array of numbers * @return int[] the sorted array / public int[] selectionSort(int[] nums) { for(int c = 0;c < nums.length;c++) { int least = nums.length-1; for(int x = c;x<nums.length;x++) // finds the smallest in the array if(nums[x] < nums[least]) least = x; int t = nums[c]; // moves to front nums[c] = nums[least]; nums[least] = t; } // then the next smallest return nums; } /* * sorts the array using insrtio (human) sort method * @param nums the array of numbers * @return int[] the sorted array */ public int[] insertionSort(int[] nums) { for(int x = 0;x<nums.length;x++) { for(int x2 = x+1; x2 < nums.length;x2++) { if(nums[x] > nums[x2])//if larger { int temp = nums[x]; // store as temp int temp2 = nums[x2]; for(int c = x2;c > x;c--) // moves everything else down { int t = nums[c-1]; nums[c] = t; //nums[c] = t; } nums[x] = temp2; // places at front } } } return nums; } } --v

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Wednesday, 02 November 2011, 20:29
JL235	You can also do: import java.util.Arrays; Arrays.sort( nums ); --v or... java.util.Arrays.sort( nums ); --v

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