Swift3.0 选择排序

选择排序
SelectionSort.swift如下：

//选择排序

var a = [6, 5, 4, 3, 2, 1]

print("array is \(a)")

var i,j,min,temp: Int

//遍历一遍数组
for i in 0..<a.count-1 {
    min = i
    
    //挨个比较其他数字
    for j in i+1..<a.count {
        print("i is \(i) and j is \(j) and min is \(min) and a[min] is \(a[min])")
        
        //看当前是否真的是最小数字
        if a[min] > a[j] {
            //a[j]更小，所以更新 min
            min = j
            
            print("change \(a[min]) and \(a[i])")
            
            let temp = a[min]
            a[min] = a[i]
            a[i] = temp
        }
    }
}

//排序结束
print("sorted array is \(a)")

Terminal运行swift SelectionSort.swift

array is [6, 5, 4, 3, 2, 1]
i is 0 and j is 1 and min is 0 and a[min] is 6
change 5 and 6
i is 0 and j is 2 and min is 1 and a[min] is 6
change 4 and 5
i is 0 and j is 3 and min is 2 and a[min] is 5
change 3 and 4
i is 0 and j is 4 and min is 3 and a[min] is 4
change 2 and 3
i is 0 and j is 5 and min is 4 and a[min] is 3
change 1 and 2
i is 1 and j is 2 and min is 1 and a[min] is 6
change 5 and 6
i is 1 and j is 3 and min is 2 and a[min] is 6
change 4 and 5
i is 1 and j is 4 and min is 3 and a[min] is 5
change 3 and 4
i is 1 and j is 5 and min is 4 and a[min] is 4
change 2 and 3
i is 2 and j is 3 and min is 2 and a[min] is 6
change 5 and 6
i is 2 and j is 4 and min is 3 and a[min] is 6
change 4 and 5
i is 2 and j is 5 and min is 4 and a[min] is 5
change 3 and 4
i is 3 and j is 4 and min is 3 and a[min] is 6
change 5 and 6
i is 3 and j is 5 and min is 4 and a[min] is 6
change 4 and 5
i is 4 and j is 5 and min is 4 and a[min] is 6
change 5 and 6
sorted array is [1, 2, 3, 4, 5, 6]