// 第一次循環(huán)i=j=0，之后每次循環(huán)j=i.
// j = ++i相當(dāng)于在每次循環(huán)的最后執(zhí)行 {i++; j = i;}
// j = i++相當(dāng)于在每次循環(huán)的最后執(zhí)行 {j = i; i++;}
for (int i = 0, j = i; i < (length - 1); j = ++i) {
    int ai = a[i + 1]; // 每次循環(huán)的目的是將下一個(gè)數(shù)排到它應(yīng)該在的位置，這里ai就是下一個(gè)數(shù)
    while (ai < a[j]) { // while循環(huán)的目的是確定j的值 和 把所有比ai大的項(xiàng)向后移一位來(lái)騰出ai的位置
        a[j + 1] = a[j]; // 把比ai大的項(xiàng)向后移一位
        if (j-- == left) { // j-- 確定j的值，也就是確定ai的位置， j 可能等于 -1
            break;
        }
    }
    a[j + 1] = ai; // j+1 就是ai的位置
}

int[] count = new int[256];

// 第一次循環(huán)：計(jì)數(shù)
for (int i = (0 - 1); ++i <= (length - 1); count[a[i] - (-128)]++);

// 第二次循環(huán)：給 < byte[] a > 賦值
// 循環(huán)結(jié)束條件以k為標(biāo)準(zhǔn)，k<=0就會(huì)停止;
// 因?yàn)閕和k沒(méi)有固定關(guān)系，所以沒(méi)有增量表達(dá)式，但在方法體中利用--i和--k進(jìn)行增量。
for (int i = 256, k = length; k > 0; ) {
    while (count[--i] == 0); // 如果計(jì)數(shù)個(gè)數(shù)為0，什么也不做，--i
    byte value = (byte) (i + (-128));
    int s = count[i];

    do {
        a[--k] = value;
    } while (--s > 0);
}

// 7等分一段的長(zhǎng)度近似值
int seventh = (length >> 3) + (length >> 6) + 1;
// 一個(gè)數(shù)組分為7段，則有五個(gè)切割點(diǎn)，如下為五個(gè)切割點(diǎn)的下標(biāo)
int e3 = (left + right) >>> 1; // The midpoint
int e2 = e3 - seventh;
int e1 = e2 - seventh;
int e4 = e3 + seventh;
int e5 = e4 + seventh;

// Sort these elements using insertion sort
if (a[e2] < a[e1]) { int t = a[e2]; a[e2] = a[e1]; a[e1] = t; }

if (a[e3] < a[e2]) { int t = a[e3]; a[e3] = a[e2]; a[e2] = t;
    if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
}
if (a[e4] < a[e3]) { int t = a[e4]; a[e4] = a[e3]; a[e3] = t;
    if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
        if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
    }
}
if (a[e5] < a[e4]) { int t = a[e5]; a[e5] = a[e4]; a[e4] = t;
    if (t < a[e3]) { a[e4] = a[e3]; a[e3] = t;
        if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
            if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
        }
    }
}

// 中心部分第一個(gè)元素的索引
int less  = left;

// 右部分第一個(gè)元素前的索引
int great = right;

// 取兩個(gè)值作為分區(qū)值
int pivot1 = a[e2];
int pivot2 = a[e4];

// 要排序的第一個(gè)和最后一個(gè)元素被移動(dòng)到以前由樞軸占據(jù)的位置。
// 當(dāng)分區(qū)完成時(shí)，軸心點(diǎn)被交換回它們的最終位置，并從隨后的排序中排除。
a[e2] = a[left];
a[e4] = a[right];

// less一開(kāi)始等于left， great一開(kāi)始等于right。
// 跳過(guò)小于或大于分割值的元素。
while (a[++less] < pivot1); // 沒(méi)有判斷第一個(gè)
while (a[--great] > pivot2); // 沒(méi)有判斷最后一個(gè)

// 循環(huán)帶outer:，`break outer;`會(huì)跳出整個(gè)循環(huán)，也就是結(jié)束整個(gè)下面的for循環(huán)。
// less不參與循環(huán)，只是一開(kāi)始給k賦值，less的變化始終是`++less`，用來(lái)交換數(shù)組中的值。
outer:
for (int k = less - 1; ++k <= great; ) {
    int ak = a[k];
    if (ak < pivot1) { // Move a[k] to left part
        a[k] = a[less];
        /*
         * Here and below we use "a[i] = b; i++;" instead
         * of "a[i++] = b;" due to performance issue.
         */
        a[less] = ak;
        ++less;
    } else if (ak > pivot2) { // Move a[k] to right part
        while (a[great] > pivot2) {
            if (great-- == k) {
                break outer;
            }
        }
        if (a[great] < pivot1) { // a[great] <= pivot2
            a[k] = a[less];
            a[less] = a[great];
            ++less;
        } else { // pivot1 <= a[great] <= pivot2
            a[k] = a[great];
        }
        /*
         * Here and below we use "a[i] = b; i--;" instead
         * of "a[i--] = b;" due to performance issue.
         */
        a[great] = ak;
        --great;
    }
}

// 循環(huán)結(jié)束，交換left和(less - 1)的值，也就是處理循環(huán)前`a[e2] = a[left];`導(dǎo)致的分區(qū)錯(cuò)誤
a[left]  = a[less  - 1]; a[less  - 1] = pivot1;
// 循環(huán)結(jié)束，交換right和(great + 1)的值，也就是處理循環(huán)前`a[e4] = a[right];`導(dǎo)致的分區(qū)錯(cuò)誤
a[right] = a[great + 1]; a[great + 1] = pivot2;

// 分為三部分后，嵌套排序第一部分和第三部分
sort(a, left, less - 2, leftmost);
sort(a, great + 2, right, false);

if (less < e1 && e5 < great) { // 剩余的中間部分超過(guò)4/7
    /*
     * Skip elements, which are equal to pivot values.
     */
    while (a[less] == pivot1) {
        ++less;
    }

    while (a[great] == pivot2) {
        --great;
    }

    outer:
    for (int k = less - 1; ++k <= great; ) {
        int ak = a[k];
        if (ak == pivot1) { // Move a[k] to left part
            a[k] = a[less];
            a[less] = ak;
            ++less;
        } else if (ak == pivot2) { // Move a[k] to right part
            while (a[great] == pivot2) {
                if (great-- == k) {
                    break outer;
                }
            }
            if (a[great] == pivot1) { // a[great] < pivot2
                a[k] = a[less];
                a[less] = pivot1;
                ++less;
            } else { // pivot1 < a[great] < pivot2
                a[k] = a[great];
            }
            a[great] = ak;
            --great;
        }
    }
}

// Sort center part recursively
sort(a, less, great, false);

// 取下標(biāo)在中間的值做一個(gè)臨時(shí)變量，該變量是中值的廉價(jià)近似值，作為分割值
int pivot = a[e3];

// less一開(kāi)始等于left， great一開(kāi)始等于right。
// 方法體內(nèi)部不斷修改great的值，使循環(huán)執(zhí)行的次數(shù)不斷的縮減，一次循環(huán)great可以減少0，可以減少1，可以減少n。
// less并不影響循環(huán)，只是作為臨時(shí)變量進(jìn)行數(shù)組中值的交換，始終小于等于k，一次循環(huán)只能加1或不加。
for (int k = less; k <= great; ++k) {
    if (a[k] == pivot) { // 如果a[k]的值等于分割值，跳過(guò)
        continue;
    }
    int ak = a[k]; // 取出a[k]值賦給臨時(shí)變量ak
    if (ak < pivot) { // Move a[k] to left part
        a[k] = a[less];
        a[less] = ak;
        ++less;
    } else { // a[k] > pivot - Move a[k] to right part
        while (a[great] > pivot) {
            --great;
        }
        if (a[great] < pivot) { // a[great] <= pivot
            a[k] = a[less];
            a[less] = a[great];
            ++less;
        } else { // a[great] == pivot
            /*
             * Even though a[great] equals to pivot, the
             * assignment a[k] = pivot may be incorrect,
             * if a[great] and pivot are floating-point
             * zeros of different signs. Therefore in float
             * and double sorting methods we have to use
             * more accurate assignment a[k] = a[great].
             */
            a[k] = pivot;
        }
        a[great] = ak;
        --great;
    }
}

// 分為三部分后，嵌套排序第一部分和第三部分
sort(a, left, less - 1, leftmost);
sort(a, great + 1, right, false);

// 合并排序中的最大運(yùn)行次數(shù)
static final int MAX_RUN_COUNT = 67;

// 合并排序中運(yùn)行的最大長(zhǎng)度
static final int MAX_RUN_LENGTH = 33;

/**
 * 長(zhǎng)度大于等于286的int數(shù)組排序
 * 
 * @param a
 *            要排序int數(shù)組
 * @param left
 *            起始下標(biāo)
 * @param right
 *            結(jié)束下標(biāo)
 * @param work
 *            null
 * @param workBase
 *            0
 * @param workLen
 *            0
 */
private static void largeSort(int[] a, int left, int right, int[] work,
			int workBase, int workLen) {

    /*
     * Index run[i] is the start of i-th run (ascending or descending
     * sequence).
     */
    int[] run = new int[MAX_RUN_COUNT + 1];
    int count = 0;
    run[0] = left;

    // Check if the array is nearly sorted
    for (int k = left; k < right; run[count] = k) {
        if (a[k] < a[k + 1]) { // ascending
            while (++k <= right && a[k - 1] <= a[k]);
        } else if (a[k] > a[k + 1]) { // descending
            while (++k <= right && a[k - 1] >= a[k]);
            for (int lo = run[count] - 1, hi = k; ++lo < --hi;) {
                int t = a[lo];
                a[lo] = a[hi];
                a[hi] = t;
            }
        } else { // equal
            for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k];) {
                if (--m == 0) {
                    sort(a, left, right, true);
                    return;
                }
            }
        }

        /*
         * The array is not highly structured, use Quicksort instead of
         * merge sort.
         */
        if (++count == MAX_RUN_COUNT) {
            sort(a, left, right, true);
            return;
        }
    }

    // Check special cases
    // Implementation note: variable "right" is increased by 1.
    if (run[count] == right++) { // The last run contains one element
        run[++count] = right;
    } else if (count == 1) { // The array is already sorted
        return;
    }

    // Determine alternation base for merge
    byte odd = 0;
    for (int n = 1; (n <<= 1) < count; odd ^= 1);

    // Use or create temporary array b for merging
    int[] b; // temp array; alternates with a
    int ao, bo; // array offsets from 'left'
    int blen = right - left; // space needed for b
    if (work == null || workLen < blen || workBase + blen > work.length) {
        work = new int[blen];
        workBase = 0;
    }
    if (odd == 0) {
        System.arraycopy(a, left, work, workBase, blen);
        b = a;
        bo = 0;
        a = work;
        ao = workBase - left;
    } else {
        b = work;
        ao = 0;
        bo = workBase - left;
    }

    // Merging
    for (int last; count > 1; count = last) {
        for (int k = (last = 0) + 2; k <= count; k += 2) {
            int hi = run[k], mi = run[k - 1];
            for (int i = run[k - 2], p = i, q = mi; i < hi; ++i) {
                if (q >= hi || p < mi && a[p + ao] <= a[q + ao]) {
					b[i + bo] = a[p++ + ao];
                } else {
				    b[i + bo] = a[q++ + ao];
				}
			}
			run[++last] = hi;
		}
		if ((count & 1) != 0) {
			for (int i = right, lo = run[count - 1]; --i >= lo;
			    b[i + bo] = a[i + ao]
			);
			run[++last] = right;
		}
		int[] t = a;
		a = b;
		b = t;
		int o = ao;
		ao = bo;
		bo = o;
	}
}