1109. 航班预订统计

这里有 n 个航班,它们分别从 1 到 n 进行编号。

有一份航班预订表 bookings ,表中第 i 条预订记录 bookings[i] = [firsti, lasti, seatsi] 意味着在从 firsti 到 lasti (包含 firsti 和 lasti )的 每个航班 上预订了 seatsi 个座位。

请你返回一个长度为 n 的数组 answer,里面的元素是每个航班预定的座位总数。

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/corporate-flight-bookings
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线段树

class Solution {
    public int[] corpFlightBookings(int[][] bookings, int n) {
        int[] ans = new int[n];
        SegmentTree segmentTree = new SegmentTree(n);
        for (int[] booking : bookings) {
            segmentTree.add(booking[0], booking[1], 1, n, booking[2], 1);
        }

        for (int i = 1; i <= n; ++i) {
            ans[i - 1] = segmentTree.query(i, i, 1, n, 1);
        }
        return ans;
    }
}

class SegmentTree {

    private int[] sum;
    private int[] add;

    public SegmentTree(int n) {
        sum = new int[n << 2 | 1];
        add = new int[n << 2 | 1];
    }

    private void pushUp(int rt) {
        sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
    }

    private void pushDown(int rt, int ln, int rn) {
        if (add[rt] != 0) {
            add[rt << 1] += add[rt];
            add[rt << 1 | 1] += add[rt];
            sum[rt << 1] += add[rt] * ln;
            sum[rt << 1 | 1] += add[rt] * rn;
            add[rt] = 0;
        }
    }

    public void add(int L, int R, int l, int r, int x, int rt) {
        if (L <= l && R >= r) {
            add[rt] += x;
            sum[rt] += (r - l + 1) * x;
            return;
        }
        int mid = (l + r) >> 1;
        pushDown(rt, mid - l + 1, r - mid);
        if (L <= mid) {
            add(L, R, l, mid, x, rt << 1);
        }
        if (mid < R) {
            add(L, R, mid + 1, r, x, rt << 1 | 1);
        }
        pushUp(rt);
    }

    public int query(int L, int R, int l, int r, int rt) {
        if (L <= l && R >= r) {
            return sum[rt];
        }
        int mid = (l + r) >> 1;
        pushDown(rt, mid - l + 1, r - mid);
        int ans = 0;
        if (L <= mid) {
            ans += query(L, R, l, mid, rt << 1);
        }
        if (mid < R) {
            ans += query(L, R, mid + 1, r, rt << 1 | 1);
        }
        pushUp(rt);
        return ans;
    }
}

差分

class Solution {
    public int[] corpFlightBookings(int[][] bookings, int n) {
        int[] ans = new int[n];
        for (int[] booking : bookings) {
            ans[booking[0] - 1] += booking[2];
            if (booking[1] < n) {
                ans[booking[1]] -= booking[2];
            }
        }
        for (int i = 1; i < n; ++i) {
            ans[i] += ans[i - 1];
        }
        return ans;
    }
}
posted @ 2022-02-16 16:03  Tianyiya  阅读(41)  评论(0)    收藏  举报