假设我们有一颗钻石（13 * 13阵列中有85个元素） 每个元素都有两个参数，a / b 我们需要对钻石进行分类，以便：
我们想要得到的是:( *是未在内存中定义的部分）
（这是一个简单的5 * 5阵列中的13个元素）
原：
* * 4/3 * *
* 1/3 2/3 4/5 *
1/2 3/1 2/5 3/6 2/7
* 2/3 1/2 5/2 *
* * 6/1 * *
排序方法：
* * 1/3 * *
* 1/2 2/3 2/5 *
1/2 2/3 4/5 3/6 2/7
* 3/1 5/2 4/3 *
* * 6/1 * *
我的算法是对所有值进行冒泡排序（根据a），然后根据b参数对每一行进行排序。这工作得很好，但动态指令将超过80000（应该是O [n ^ 2]）。泡沫排序在MIPS中很难搞，因为你需要＆gt; 10 instr。每次检查（泡泡）。
我想知道是否有更好的算法可以减少14000以下的动态指令.BTW只使用＆lt; = 15寄存器和＆lt; = 66静态指令（66行代码）。
findnum: addi $27, $27, 4
lw $21, 0($27)
beq $21, $0, findnum # find the next nonzero number
bsort: lw $20, 0($25) # load first number
slt $12, $21, $20 # if second < first $12=1
bne $12, $15, back # $15 is constant 1
sw $21, 0($25)
sw $20, 0($27)
back: addi $25, $27, 0 # now second number is first number
bne $27, $26, findnum # $27 not the end then continue
loop: addi $25, $01, 24
addi $27, $25, 0 # reinitialize the number
addi $10, $10, 1 # i++
bne $10, $11, findnum # if i not equal to 85 jump back
这是我的代码的核心部分。 25美元和27美元被初始化为数组第一个值的地址。 $ 26是最后一个值的地址。
非常感谢！
答案 0 :(得分：0)
允许您证明2D结构的一种方法是转换您的值，将它们填充到一维数组中，对它们进行排序，将它们填充回2D并将它们转换回来。
因此，通过简化示例，此4/3将成为此变换的数字：403：
n(x/y) = 100*x + y
将2D数组填充到1D数组是一个简单的（嵌套的）for循环，然后再返回。
[403, 103, 203, 405 ...]
然后，一旦你在一维数组中有这样的数字，只需对它进行排序，例如通过Quicksort。
要将1D复制回2D，您应该有一个规则，哪个数字属于哪一行。数字的转换将是：
(x/y) = ( floor(n/100) / n%100 )
使用一般标准对2D结构进行一般排序可能并不那么容易。但在这种情况下，这种方法将起作用。它已被用于其他场合：
答案 1 :(得分：0)
SortDiamond: addi $01, $00, Array # set memory base
swi 521 # create sort diamond and update memory
addi $06, $00, 6 # constant 6
addi $07, $00, 12 # constant 12
addi $12, $00, 0 # counter inner
addi $21, $00, 0 # counter in bubble
addi $22, $00, 0 # bs counter
bigloop: addi $20, $01, 72 # find the address first element to sort(row)
addi $26, $20, 0
addi $10, $00, 1 # i starts from 1
addi $11, $00, 2 # sort times
addi $23, $00, 2
bsrow: lw $17, 0($26)
lw $18, 4($26)
andi $27, $17, 0x003F
andi $28, $18, 0x003F
slt $19, $28, $27
beq $19, $00, go
sw $17, 4($26)
sw $18, 0($26)
addi $22, $00, 1 # this is the indicator of bubble sort
go: addi $26, $26, 4
addi $21, $21, 1
bne $21, $23, bsrow
loopi: addi $12, $12, 1
addi $26, $20, 0
addi $23, $23, -1
addi $21, $00, 0
beq $22, $00, loopj # jump if it doesn't do bubble sort in one loop
bne $12, $11, bsrow
loopj: addi $10, $10, 1 # i++
addi $22, $00, 0 # reinitialize counter
slti $19, $10, 7 # find next row
beq $19, $00, great # if i<7 next address is 48+ current
sll $11, $10, 1
addi $20, $20, 48
j less
great: sub $11, $07, $10
sll $11, $11, 1
addi $20, $20, 56
less: addi $23, $11, 0
addi $26, $20, 0
addi $12, $00, 0
bne $10, $07, bsrow
addi $20, $01, 264
addi $26, $20, 0
addi $10, $00, 1
addi $23, $00, 2
addi $11, $00, 2
bscol: lw $17, 0($26)
lw $18, 52($26)
slt $19, $18, $17
beq $19, $00, go1
sw $17, 52($26)
sw $18, 0($26)
addi $22, $00, 1
go1: addi $26, $26, 52
addi $21, $21, 1
bne $21, $23, bscol
loopi1: addi $12, $12, 1
addi $26, $20, 0
addi $23, $23, -1
addi $21, $00, 0
beq $22, $00, loopj1
bne $12, $11, bscol
loopj1: addi $10, $10, 1
addi $22, $00, 0
slti $19, $10, 7
beq $19, $00, great1
sll $11, $10, 1
addi $20, $20, -48
j less1
great1: sub $11, $07, $10
sll $11, $11, 1
addi $20, $20, 56
less1: addi $23, $11, 0
addi $26, $20, 0
addi $12, $00, 0
bne $10, $07, bscol
swi 522
bne $02, $00, bigloop
swi 523 # redisplay diamond
jr $31 # return to caller