//需要 Swift 3.0 環(huán)境

//從小到大排序

var a = [6, 5, 4, 3, 2, 1]
print("a is \(a)")

//從第二個開始遍歷到最后一個
for j in 1...a.count-1 {
    //保留第 j 個的值，因?yàn)楹竺娴牟僮鲿薷臄?shù)組
    let n = a[j]
    print("now moving by \(n)")
    
    //記錄 j-1
    var i = j - 1
    //當(dāng)前面還有數(shù)，并且前面的數(shù)比保留值大，則前后數(shù)交換位置
    while i >= 0 && a[i] > n {
        print("change \(a[i+1]) and \(a[i])")
        
        let temp = a[i+1]
        a[i+1] = a[i]
        a[i] = temp
        
        //繼續(xù)判斷前面的數(shù)
        i = i - 1
        
        print("j = \(j) and i = \(i) and a = \(a)")
    }
    print("")
}

//排序完成
print("sorted a is \(a)")
}

之后在Terminal運(yùn)行swift Numbers.swift即可看到輸出

a is [6, 5, 4, 3, 2, 1]
now moving by 5
change 5 and 6
j = 1 and i = -1 and a = [5, 6, 4, 3, 2, 1]

now moving by 4
change 4 and 6
j = 2 and i = 0 and a = [5, 4, 6, 3, 2, 1]
change 4 and 5
j = 2 and i = -1 and a = [4, 5, 6, 3, 2, 1]

now moving by 3
change 3 and 6
j = 3 and i = 1 and a = [4, 5, 3, 6, 2, 1]
change 3 and 5
j = 3 and i = 0 and a = [4, 3, 5, 6, 2, 1]
change 3 and 4
j = 3 and i = -1 and a = [3, 4, 5, 6, 2, 1]

now moving by 2
change 2 and 6
j = 4 and i = 2 and a = [3, 4, 5, 2, 6, 1]
change 2 and 5
j = 4 and i = 1 and a = [3, 4, 2, 5, 6, 1]
change 2 and 4
j = 4 and i = 0 and a = [3, 2, 4, 5, 6, 1]
change 2 and 3
j = 4 and i = -1 and a = [2, 3, 4, 5, 6, 1]

now moving by 1
change 1 and 6
j = 5 and i = 3 and a = [2, 3, 4, 5, 1, 6]
change 1 and 5
j = 5 and i = 2 and a = [2, 3, 4, 1, 5, 6]
change 1 and 4
j = 5 and i = 1 and a = [2, 3, 1, 4, 5, 6]
change 1 and 3
j = 5 and i = 0 and a = [2, 1, 3, 4, 5, 6]
change 1 and 2
j = 5 and i = -1 and a = [1, 2, 3, 4, 5, 6]

sorted a is [1, 2, 3, 4, 5, 6]

如果我們想要從大往小排序，只需要修改while i >= 0 && a[i] > n {...}為while i >= 0 && a[i] < n {...}就可以了。