开始Java：排序方法说明

这是一个排序方法的示例：

    int sorting = 0;
    for(int i = 0; i<myArray.length-1; i++){
        for(int s = i+1; s<myArray.length; s++){
            if(myArray[i] > myArray[s]){
                sorting = myArray[i];
                myArray[i] = myArray[s]; 
                myArray[s] = sorting; 
            }
        }
    }

int排序=0；
对于（inti=0；i也许这个解释有帮助

循环不变量：直到i-1的元素被排序，从i到所有
大于或等于

/“将对小于等于-1（=无）的元素进行排序，从0开始，所有元素都大于或等于”
for（int i=0；imyArray[s]）{//如果有一个较小的
//在i和s处交换元素：
int oldValueAtI=myArray[i]；
myArray[i]=myArray[s]；
myArray[s]=oldValueAtI；
}
}
//对于s>i，myArray[i]是最小的
//到i的元素被排序，从i+1开始都大于或等于
}
//对长度小于等于s.length-1的元素进行排序，
//从s.length（=无）开始，所有长度都大于或等于
//所有元素都已排序


因此，您甚至可以简化工作，减少交换：

for (int i = 0; i < myArray.length-1; i++){
    // Search the smallest in the rest:
    int smallestS = i;
    for (int s = i+1; s < myArray.length; s++) {
        if (myArray[s] < myArray[smallestS]) {
            smallestS = s;
        }
    }
    // Swap the elements at i and smallestS
    if (smallestS != i) { // Check not needed
        int oldValueAtI = myArray[i];
        myArray[i] = myArray[smallestS]; 
        myArray[smallestS] = oldValueAtI;
    } 
}

for（int i=0；i
从这里看起来像是冒泡排序。非常低效的排序。任何进一步的细化都会复制算法书或维基百科，真的。是的，我现在是初学者，所以我并不真正关心事情的效率。只有它的工作方式，我才知道它为什么会这样。有人能解释吗？Makoto是正确的是你可以通过简单的谷歌搜索轻松找到的东西
// "The elements upto -1 (=none) are sorted, from 0 are all greater or equal"
for (int i = 0; i < myArray.length-1; i++){

    // The elements upto  i-1 are sorted, from i are all greater or equal
    for (int s = i+1; s < myArray.length; s++) { // For the elements after i
        if (myArray[i] > myArray[s]) { // If there is a smaller one
            // Swap the elements at i and s:
            int oldValueAtI = myArray[i];
            myArray[i] = myArray[s]; 
            myArray[s] = oldValueAtI; 
        }
    }
    // myArray[i] is smallest for s > i
    // The elements upto  i are sorted, from i+1 are all greater or equal
}
// The elements upto s.length - 1 are sorted,
// from s.length (=none) are all greater or equal

// All elements are sorted

for (int i = 0; i < myArray.length-1; i++){
    // Search the smallest in the rest:
    int smallestS = i;
    for (int s = i+1; s < myArray.length; s++) {
        if (myArray[s] < myArray[smallestS]) {
            smallestS = s;
        }
    }
    // Swap the elements at i and smallestS
    if (smallestS != i) { // Check not needed
        int oldValueAtI = myArray[i];
        myArray[i] = myArray[smallestS]; 
        myArray[smallestS] = oldValueAtI;
    } 
}