剑指 Offer 59 - I. 滑动窗口的最大值
解法一:暴力模拟
class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
if (nums.length == 0) return new int[0];
int len = nums.length - k + 1;
List<Integer> list = new ArrayList<>();
int[] ret = new int[len];
for (int i = 0; i < len; ++i) {
int max = Integer.MIN_VALUE;
for (int j = i; j < i + k; ++j) {
max = Math.max(nums[j], max);
}
list.add(max);
}
for (int i = 0; i < len; ++i) {
ret[i] = list.get(i);
}
return ret;
}
}
解法二:单调队列(优化版)
class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
if (nums.length == 0 || k == 0) return new int[0];
Deque<Integer> deque = new LinkedList<>();
int[] ret = new int[nums.length - k + 1];
int index = 0;
//未形成窗口区间
for (int i = 0; i < k; ++i) {
//队列不为空时,当前值与队列尾部值比较,如果大于,删除队列尾部值
//一直循环删除到队列中的值都大于当前值,或者删到队列为空
while (!deque.isEmpty() && nums[i] > deque.peekLast()) {
deque.removeLast();
}
//执行完上面的循环后,队列中要么为空,要么值都比当前值大,然后就把当前值添加到队列中
deque.addLast(nums[i]);
}
//窗口区间刚形成后,把队列首位值添加到队列中
ret[index++] = deque.peekFirst();
//窗口区间形成
for (int i = k; i < nums.length; ++i) {
//i-k是已经在区间外了,如果首位等于nums[i-k],那么说明此时首位值已经不再区间内了,需要删除
if (deque.peekFirst() == nums[i - k]) {
deque.removeFirst();
}
//删除队列中比当前值小的值
while (!deque.isEmpty() && deque.peekLast() < nums[i]) {
deque.removeLast();
}
//把当前值添加到队列中
deque.addLast(nums[i]);
//把队列的首位值添加到arr数组中
ret[index++] = deque.peekFirst();
}
return ret;
}
}
剑指 Offer 59 - II. 队列的最大值
解法:滑动窗口
class MaxQueue {
Deque<Integer> ret, max;
public MaxQueue() {
ret = new LinkedList<Integer>();
max = new LinkedList<Integer>();
}
public int max_value() {
if (max.isEmpty()) return -1;
return max.peekFirst();
}
public void push_back(int value) {
ret.addLast(value);
while (!max.isEmpty() && max.peekLast() < value) max.removeLast();
max.addLast(value);
}
public int pop_front() {
if (ret.isEmpty()) return -1;
int temp = ret.peekFirst();
if (temp == max.peekFirst()) max.removeFirst();
ret.removeFirst();
return temp;
}
}