普通插入排序与成对插入排序

for (int i = left, j = i; i < right; j = ++i)

{
                    int ai = a[i + 1];
                    while (ai < a[j]) {
                        a[j + 1] = a[j];
                        if (j-- == left) {
                            break;
                        }
                    }
                    a[j + 1] = ai;
                }

成对插入排序，提高了插入排序的性能，同时可以插入两个数值

do {
                    if (left >= right) {
                        return;
                    }
                } while (a[++left] >= a[left - 1]);

                /*
                 * Every element from adjoining part plays the role
                 * of sentinel, therefore this allows us to avoid the
                 * left range check on each iteration. Moreover, we use
                 * the more optimized algorithm, so called pair insertion
                 * sort, which is faster (in the context of Quicksort)
                 * than traditional implementation of insertion sort.

    ​    ​    ​    ​ * 具体执行过程：上面的do-while循环已经排好的最前面的数据

    ​    ​    ​    ​*（1）将要插入的数据，第一个值赋值a1,第二个值赋值a2,

    ​    ​    ​    ​*(2)然后判断a1与a2的大小，使a1要大于a2

    ​    ​    ​    ​*(3)接下来，首先是插入大的数值a1，将a1与k之前的数字一一比较，直到数值小于a1为止，把a1插入到合适的位置，注意：这里的相隔距离为2

    ​            *(4)接下来，插入小的数值a2,将a2与此时k之前的数字一一比较，直到数值小于a2为止，将a2插入到合适的位置，注意：这里的相隔距离为1

                *(5)最后把最后一个没有遍历到的数据插入到合适位置

                 */
                for (int k = left; ++left <= right; k = ++left) {
                    int a1 = a[k], a2 = a[left];

                    if (a1 < a2) {
                        a2 = a1; a1 = a[left];
                    }
                    while (a1 < a[--k]) {
                        a[k + 2] = a[k];
                    }
                    a[++k + 1] = a1;

                    while (a2 < a[--k]) {
                        a[k + 1] = a[k];
                    }
                    a[k + 1] = a2;
                }
                int last = a[right];

                while (last < a[--right]) {
                    a[right + 1] = a[right];
                }
                a[right + 1] = last;
            }
            return;