单调栈、排序哈希映射 2 算法，slice() / Arrays.copyOfRange / Arrays.Copy / append([]int{}, ar...) / memcpy 拷贝数组，求解《768. 最多能完成排序的块 II》
例题
768. 最多能完成排序的块 II
这个问题和“最多能完成排序的块”相似，但给定数组中的元素可以重复，输入数组最大长度为2000，其中的元素最大为1e8。
arr是一个可能包含重复元素的整数数组，我们将这个数组分割成几个“块”，并将这些块分别进行排序。之后再连接起来，使得连接的结果和按升序排序后的原数组相同。
我们最多能将数组分成多少块？
示例 1:
输入: arr = [5,4,3,2,1]
输出: 1
解释:
将数组分成2块或者更多块，都无法得到所需的结果。
例如，分成 [5, 4], [3, 2, 1] 的结果是 [4, 5, 1, 2, 3]，这不是有序的数组。
示例 2:
输入: arr = [2,1,3,4,4]
输出: 4
解释:
我们可以把它分成两块，例如 [2, 1], [3, 4, 4]。
然而，分成 [2, 1], [3], [4], [4] 可以得到最多的块数。
注意:
arr的长度在[1, 2000]之间。
arr[i]的大小在[0, 1e8]之间。
答案
单调栈
var maxChunksToSorted = function(arr) {
const n = arr.length, monotoneStack = []
for (let i = 0; i < n; i++) {
if (monotoneStack.length === 0 || arr[i] >= monotoneStack[monotoneStack.length - 1]) {
monotoneStack.push(arr[i])
} else {
const maxNum = monotoneStack.pop()
while (monotoneStack.length > 0 && arr[i] < monotoneStack[monotoneStack.length - 1]) {
monotoneStack.pop()
}
monotoneStack.push(maxNum)
}
}
return monotoneStack.length
};function maxChunksToSorted(arr: number[]): number {
const n = arr.length, monotoneStack = []
for (let i = 0; i < n; i++) {
if (monotoneStack.length === 0 || monotoneStack[monotoneStack.length - 1] <= arr[i]) {
monotoneStack.push(arr[i])
} else {
const maxNum = monotoneStack.pop()
while (monotoneStack.length && monotoneStack[monotoneStack.length - 1] > arr[i]) monotoneStack.pop()
monotoneStack.push(maxNum)
}
}
return monotoneStack.length
};class Solution {
function maxChunksToSorted($arr) {
$monotoneStack = [];
foreach($arr as $num) {
if (count($monotoneStack) === 0 || end($monotoneStack) <= $num) {
$monotoneStack []= $num;
} else {
$maxNum = array_pop($monotoneStack);
while (count($monotoneStack) > 0 && end($monotoneStack) > $num) {
array_pop($monotoneStack);
}
$monotoneStack []= $maxNum;
}
}
return count($monotoneStack);
}
}func maxChunksToSorted(arr []int) int {
monotoneStack := []int{}
for _, num := range arr {
if len(monotoneStack) == 0 || monotoneStack[len(monotoneStack) - 1] <= num {
monotoneStack = append(monotoneStack, num)
} else {
maxNum := monotoneStack[len(monotoneStack) - 1]
monotoneStack = monotoneStack[:len(monotoneStack) - 1]
for len(monotoneStack) > 0 && monotoneStack[len(monotoneStack) - 1] > num {
monotoneStack = monotoneStack[:len(monotoneStack) - 1]
}
monotoneStack = append(monotoneStack, maxNum)
}
}
return len(monotoneStack)
}class Solution {
public int maxChunksToSorted(int[] arr) {
Deque<Integer> monotoneStack = new ArrayDeque<Integer>();
for (int num : arr) {
if (monotoneStack.isEmpty() || monotoneStack.peek() <= num) {
monotoneStack.push(num);
} else {
int maxNum = monotoneStack.pop();
while (monotoneStack.isEmpty() == false && monotoneStack.peek() > num) monotoneStack.pop();
monotoneStack.push(maxNum);
}
}
return monotoneStack.size();
}
}public class Solution {
public int MaxChunksToSorted(int[] arr) {
Stack<int> monotoneStack = new Stack<int>();
foreach (int num in arr) {
if (monotoneStack.Count == 0 || monotoneStack.Peek() <= num) {
monotoneStack.Push(num);
} else {
int maxNum = monotoneStack.Pop();
while (monotoneStack.Count > 0 && monotoneStack.Peek() > num) monotoneStack.Pop();
monotoneStack.Push(maxNum);
}
}
return monotoneStack.Count;
}
}int maxChunksToSorted(int* arr, int arrSize){
int* monotoneStack = malloc(sizeof(int) * 2000);
int p = 0;
for (int i = 0; i < arrSize; i++) {
if (p == 0 || monotoneStack[p - 1] <= arr[i]) monotoneStack[p++] = arr[i];
else {
int maxNum = monotoneStack[--p];
while (p > 0 && monotoneStack[p - 1] > arr[i]) p--;
monotoneStack[p++] = maxNum;
}
}
return p;
}class Solution {
public:
int maxChunksToSorted(vector<int>& arr) {
stack<int> monotoneStack;
for (int num : arr) {
if (monotoneStack.empty() || monotoneStack.top() <= num) monotoneStack.push(num);
else {
int maxNum = monotoneStack.top();
monotoneStack.pop();
while (monotoneStack.empty() == false && monotoneStack.top() > num) monotoneStack.pop();
monotoneStack.push(maxNum);
}
}
return monotoneStack.size();
}
};class Solution:
def maxChunksToSorted(self, arr: List[int]) -> int:
monotoneStack = []
for num in arr:
if len(monotoneStack) == 0 or monotoneStack[-1] <= num: monotoneStack.append(num)
else:
maxNum = monotoneStack.pop()
while len(monotoneStack) > 0 and monotoneStack[-1] > num: monotoneStack.pop()
monotoneStack.append(maxNum)
return len(monotoneStack)排序 · 哈希映射
var maxChunksToSorted = function(arr) {
const n = arr.length, h = new Map, a = arr.slice().sort((a, b) => a - b)
let r = 0
for (let i = 0; i < n; i++) {
add(h, arr[i], 1)
add(h, a[i], -1)
if (h.size === 0) r++
}
return r
};
function add(h, k, v) {
v += h.get(k) || 0
if (v === 0) h.delete(k)
else h.set(k, v)
}function maxChunksToSorted(arr: number[]): number {
const n = arr.length, h = {c:Object.create(null), size: 0}, a = arr.slice().sort((a, b) => a - b)
let r = 0
for (let i = 0; i < n; i++) {
add(h, arr[i], 1)
add(h, a[i], -1)
if (h.size === 0) r++
}
return r
};
function add(h: {c: Object, size: number}, k: number, v: number) {
if (h.c[k] === void 0) {
h.size++
h.c[k] = v
} else {
h.c[k] += v
if (h.c[k] === 0) {
delete h.c[k]
h.size--
}
}
}class Solution {
private $h = [];
function maxChunksToSorted($arr) {
$a = $arr;
usort($a, function($a, $b) {
return $a > $b;
});
$n = count($arr);
$r = 0;
for ($i = 0; $i < $n; $i++) {
$this->add($a[$i], 1);
$this->add($arr[$i], -1);
if (count($this->h) == 0) $r++;
}
return $r;
}
function add($k, $v) {
$v += $this->h[$k] ?? 0;
if ($v === 0) unset($this->h[$k]);
else $this->h[$k] = $v;
}
}func maxChunksToSorted(arr []int) int {
h, a, r := map[int]int{}, append([]int{}, arr...), 0
sort.Ints(a)
for i, num := range arr {
add(&h, num, +1)
add(&h, a[i], -1)
if len(h) == 0 {
r++
}
}
return r
}
func add(h *map[int]int, k int, v int) {
(*h)[k] += v
if (*h)[k] == 0 {
delete(*h, k)
}
}class Solution {
public int maxChunksToSorted(int[] arr) {
Map<Integer, Integer> h = new HashMap<Integer, Integer>();
int n = arr.length, r = 0;
int[] a = Arrays.copyOfRange(arr, 0, n);
Arrays.sort(a);
for (int i = 0; i < n; i++) {
add(h, a[i], 1);
add(h, arr[i], -1);
if (h.isEmpty()) r++;
}
return r;
}
private void add(Map<Integer, Integer> h, int k, int v) {
if (h.containsKey(k)) {
h.put(k, h.get(k) + v);
if (h.get(k) == 0) h.remove(k);
} else h.put(k, v);
}
}public class Solution {
public int MaxChunksToSorted(int[] arr) {
int n = arr.Length, r = 0;
int[] a = new int[n];
Array.Copy(arr, 0, a, 0, n);
Array.Sort(a);
Dictionary<int, int> h = new Dictionary<int, int>();
for(int i = 0; i < n; i++) {
Add(h, arr[i], 1);
Add(h, a[i], -1);
if (h.Count == 0) r++;
}
return r;
}
private void Add(Dictionary<int, int> h, int k, int v) {
if (h.ContainsKey(k)) {
h[k] += v;
if (h[k] == 0) h.Remove(k);
} else {
h.Add(k, v);
}
}
}typedef struct {
int k;
int v;
UT_hash_handle hh;
} HashItem;
static inline int cmp(const void *pa, const void *pb) {
return *(int*)pa - *(int*)pb;
}
void add(HashItem* *h, int k, int v) {
HashItem *pEntry = NULL;
HASH_FIND_INT(*h, &k, pEntry);
if (pEntry == NULL) {
pEntry = malloc(sizeof(HashItem));
pEntry->k = k;
pEntry->v = 0;
HASH_ADD_INT(*h, k, pEntry);
}
pEntry->v += v;
if (pEntry->v == 0) {
HASH_DEL(*h, pEntry);
free(pEntry);
}
}
int maxChunksToSorted(int* arr, int arrSize){
HashItem *h = NULL;
int* a = malloc(sizeof(int) * arrSize);
memcpy(a, arr, sizeof(int) * arrSize);
qsort(a, arrSize, sizeof(int), cmp);
int r = 0;
for (int i = 0; i < arrSize; i++) {
add(&h, a[i], 1);
add(&h, arr[i], -1);
if (HASH_COUNT(h) == 0) r++;
}
return r;
}class Solution {
public:
int maxChunksToSorted(vector<int>& arr) {
unordered_map<int, int> h;
int n = arr.size(), r = 0;
vector<int> a = arr;
sort(a.begin(), a.end());
for (int i = 0; i < n; i++) {
add(&h, a[i], 1);
add(&h, arr[i], -1);
if (h.size() == 0) r++;
}
return r;
}
void add(unordered_map<int, int> *h, int k, int v) {
(*h)[k] += v;
if ((*h)[k] == 0) (*h).erase(k);
}
};class Solution:
def maxChunksToSorted(self, arr: List[int]) -> int:
h, r, a = {}, 0, sorted(arr) # 字典
for i, num in enumerate(arr):
self.add(h, num, 1)
self.add(h, a[i], -1)
if len(h) == 0: r += 1
return r
def add(self, h: Dict[int, int], k: int, v: int):
h[k] = h.get(k, 0) + v
if h[k] == 0: del h[k]class Solution:
def maxChunksToSorted(self, arr: List[int]) -> int:
h, r, a = Counter(), 0, sorted(arr) # 可计数哈希对象
for i, num in enumerate(arr):
self.add(h, num, 1)
self.add(h, a[i], -1)
if len(h) == 0: r += 1
return r
def add(self, h: Counter, k: int, v: int):
h[k] += v
if h[k] == 0: del h[k]