题目： 88. 合并两个有序数组
题目说明
给你两个有序整数数组 nums1 和 nums2，请你将 nums2 合并到 nums1 中，使 nums1 成为一个有序数组。
说明:
初始化 nums1 和 nums2 的元素数量分别为 m 和 n 。
你可以假设 nums1 有足够的空间（空间大小大于或等于 m + n）来保存 nums2 中的元素。
示例:
输入:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6], n = 3
输出: [1,2,2,3,5,6]
解题思路：
- 合并两个数组
- 对合并的数组进行排序
源码：
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
System.arraycopy(nums2,0,nums1,m,n);
Arrays.sort(nums1);
}
}
System.arraycopy（）的源码见下
/**
* @param src the source array.
* @param srcPos starting position in the source array.
* @param dest the destination array.
* @param destPos starting position in the destination data.
* @param length the number of array elements to be copied.
* @throws IndexOutOfBoundsException if copying would cause
* access of data outside array bounds.
* @throws ArrayStoreException if an element in the {@code src}
* array could not be stored into the {@code dest} array
* because of a type mismatch.
* @throws NullPointerException if either {@code src} or
* {@code dest} is {@code null}.
*/
@HotSpotIntrinsicCandidate
public static native void arraycopy(Object src, int srcPos,
Object dest, int destPos,
int length);
Arrays.sort() 源码
/**
* Sorts the specified array into ascending numerical order.
*
* Implementation note: The sorting algorithm is a Dual-Pivot Quicksort
* by Vladimir Yaroslavskiy, Jon Bentley, and Joshua Bloch. This algorithm
* offers O(n log(n)) performance on many data sets that cause other
* quicksorts to degrade to quadratic performance, and is typically
* faster than traditional (one-pivot) Quicksort implementations.
*
* @param a the array to be sorted
*/
public static void sort(int[] a) {
DualPivotQuicksort.sort(a, 0, a.length - 1, null, 0, 0);
}
Class DualPivotQuicksort {
// ....省略..
/**
* The maximum number of runs in merge sort.
*/
private static final int MAX_RUN_COUNT = 67;
/**
* If the length of an array to be sorted is less than this
* constant, Quicksort is used in preference to merge sort.
*/
private static final int QUICKSORT_THRESHOLD = 286;
/**
* Sorts the specified range of the array using the given
* workspace array slice if possible for merging
*
* @param a the array to be sorted
* @param left the index of the first element, inclusive, to be sorted
* @param right the index of the last element, inclusive, to be sorted
* @param work a workspace array (slice)
* @param workBase origin of usable space in work array
* @param workLen usable size of work array
*/
static void sort(int[] a, int left, int right,
int[] work, int workBase, int workLen) {
// Use Quicksort on small arrays
if (right - left < QUICKSORT_THRESHOLD) {
sort(a, left, right, true);
return;
}
/*
* 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) {
// Equal items in the beginning of the sequence
while (k < right && a[k] == a[k + 1])
k++;
if (k == right) break; // Sequence finishes with equal items
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]);
// Transform into an ascending sequence
for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {
int t = a[lo]; a[lo] = a[hi]; a[hi] = t;
}
}
// Merge a transformed descending sequence followed by an
// ascending sequence
if (run[count] > left && a[run[count]] >= a[run[count] - 1]) {
count--;
}
/*
* The array is not highly structured,
* use Quicksort instead of merge sort.
*/
if (++count == MAX_RUN_COUNT) {
sort(a, left, right, true);
return;
}
}
// These invariants should hold true:
// run[0] = 0
// run[] = right + 1; (terminator)
if (count == 0) {
// A single equal run
return;
} else if (count == 1 && run[count] > right) {
// Either a single ascending or a transformed descending run.
// Always check that a final run is a proper terminator, otherwise
// we have an unterminated trailing run, to handle downstream.
return;
}
right++;
if (run[count] < right) {
// Corner case: the final run is not a terminator. This may happen
// if a final run is an equals run, or there is a single-element run
// at the end. Fix up by adding a proper terminator at the end.
// Note that we terminate with (right + 1), incremented earlier.
run[++count] = right;
}
// 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;
}
}
// ....省略..
}
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