LC.P729[我的日程安排表I]

方法一:模拟

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class MyCalendar {

List<int[]> booked;

public MyCalendar() {
booked = new ArrayList<int[]>();
}

public boolean book(int start, int end) {
for (int[] arr : booked) {
int left = arr[0], right = arr[1];
if (left < end && start < right) {
return false;
}
}
booked.add(new int[]{start, end});
return true;
}
}

/**
* Your MyCalendar object will be instantiated and called as such:
* MyCalendar obj = new MyCalendar();
* boolean param_1 = obj.book(start,end);
*/
  • 时间复杂度:$O(n^2)$
  • 空间复杂度:$O(n)$

方法二:线段树(动态开点)

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class MyCalendar {

public MyCalendar() {

}

public boolean book(int start, int end) {
// 先查询该区间是否为 0
if (query(root, 0, N, start, end - 1) != 0) return false;
// 更新该区间
update(root, 0, N, start, end - 1, 1);
return true;
}

private int N = (int) 1e9;
private Node root = new Node();

public void update(Node node, int start, int end, int l, int r, int val) {
if (l <= start && end <= r) {
node.val += val;
node.add += val;
return ;
}
pushDown(node);
int mid = (start + end) >> 1;
if (l <= mid) update(node.left, start, mid, l, r, val);
if (r > mid) update(node.right, mid + 1, end, l, r, val);
pushUp(node);
}

public int query(Node node, int start, int end, int l, int r) {
if (l <= start && end <= r) return node.val;
pushDown(node);
int mid = (start + end) >> 1, ans = 0;
if (l <= mid) ans = query(node.left, start, mid, l, r);
if (r > mid) ans = Math.max(ans, query(node.right, mid + 1, end, l, r));
return ans;
}

private void pushUp(Node node) {
// 每个节点存的是当前区间的最大值
node.val = Math.max(node.left.val, node.right.val);
}

private void pushDown(Node node) {
if (node.left == null) node.left = new Node();
if (node.right == null) node.right = new Node();
if (node.add == 0) return;
node.left.val += node.add;
node.right.val += node.add;
node.left.add += node.add;
node.right.add += node.add;
node.add = 0;
}

private static class Node {
// 左右孩子节点
Node left, right;
// 当前节点值,以及懒惰标记的值
int val, add;
}
}

/**
* Your MyCalendar object will be instantiated and called as such:
* MyCalendar obj = new MyCalendar();
* boolean param_1 = obj.book(start,end);
*/
  • 时间复杂度:$O(logn)$
  • 空间复杂度:$O(mlogn)$,$m$为查询次数