Selection Sort (java code implementation)

Algorithm Description:

• In a disordered array of length N, the n-1 first pass find the minimum number and the first number of the exchange.
• Traversing the second time n-2 from the number of the next
  number, find the minimum number and the second number of the exchange.
• Repeat these steps until the n-1 traversal minimum number n-1 and the number of exchange, the sort is complete.
  

For example

Change in said first state of each step, the details behind the introduction, the conventional random array [624159]

Find the smallest number 1 first trip into the far front (with the first digital switching)

Before the exchange: | 6 | 2 | 4 | 1 | 5 | 9 |

After the exchange: | 1 | 2 | 4 | 6 | 5 | 9 |

A second minimum number of times to find the remaining number [24659] Lane 2, the current exchanged with the first digit of the array, not the actual exchange, already at the first place

Before switching: | 1 | 2 | 4 | 6 | 5 | 9 |

After the exchange: | 1 | 2 | 4 | 6 | 5 | 9 |

Continue to find the remaining third trip [4659] minimum number 4 in the figures, there is no actual exchange, 4 need to be exchanged first position

The fourth times to find the minimum number of 5, and the first digit from the remaining [659] switched in position 6

Before switching: | 1 | 2 | 4 | 6 | 5 | 9 |

After the exchange: | 1 | 2 | 4 | 5 | 6 | 9 |

The fifth trip to find from the remaining [69] where the minimum number of 6 and found it to stay in the correct position, there is no exchange

Sorted output the correct result [124569]

The first details of the trip to find the minimum number of 1

Current array is | 6 | 2 | 4 | 1 | 5 | 9 |

6 first taken out, it played the minimum number of

The current minimum number 6 and number one by the other, it is found more decimal exchanged roles

6 compared with the current minimum number of 2, was found more decimal, switch roles, the minimum number is 2 at this time, compared with the next two remaining digits

2 compared with the current minimum number of 4, fixed

The current minimum number 2 and 1, it is found more decimal, switch roles, at this time is a minimum number, then a comparison with the remaining figures

Compared with the current minimum number of 51, fixed

The current minimum number of comparison 1 and 9, fixed, reach the end

The current minimum number of exchange 1 to the current position of the first number, as shown in FIG.

Before the exchange: | 6 | 2 | 4 | 1 | 5 | 9 |

After the exchange: | 1 | 2 | 4 | 6 | 5 | 9 |

Sorting trip completion, the remaining steps are similar

package sort;

public class 选择排序 {

	/**
	 * @param args
	 */
	public static void main(String[] args) {

		int []arr={101,34,119,1};
		selectSort(arr);
		for (int i = 0; i < arr.length; i++) {
			System.out.println(arr[i]);
		}
	}
	public static void  selectSort(int[] arr) {
		for (int i = 0; i < arr.length-1; i++) {//此时的i < arr.length-1;与i < arr.length无大区别，当循环三次时，最后一个数自然出现
			int minIndex=i;
			int min=arr[i];
			for (int j = i+1; j < arr.length; j++) {//此时的j=i+1 与j=i没有大区别，第一个是假定的第一位最小值 不与本身比较；第二个 是与本身比较
				if(min>arr[j]){
					min=arr[j];
					minIndex=j;
				}		
			}
			
			if(minIndex !=i){
				/*arr[minIndex]=arr[i];
				arr[i]=min;
				*/
				int temp = arr[i];
                arr[i] = arr[minIndex];
                arr[minIndex] = temp;
			}
		}
		
	
	}

}