LeetCode 406

class Solution {
    int[] tr;
    int n;
    
    private int lowbit(int x) {
        return x & -x;
    }
    
    private void add(int x, int v) {
        for (int i = x; i <= n; i += lowbit(i)) {
            tr[i] += v;
        }
    }
    
    // 计算[1,x]的和
    private int query(int x) {
        int res = 0;
        for (int i = x; i > 0; i -= lowbit(i)) {
            res += tr[i];
        }
        return res;
    }
    
    public int[][] reconstructQueue(int[][] people) {
        n = people.length;
        tr = new int[n + 1];
        
        Arrays.sort(people, (o1, o2) -> o1[0] != o2[0] ? o1[0] - o2[0] : o2[1] - o1[1]);
        
        int[][] res = new int[n][2];
    
        for (int[] p : people) {
            int h = p[0], k = p[1];
            int l = 1, r = n;
            while (l < r) {
                int mid = l + r >> 1;
                // mid - query(mid) , [1,mid]中0的数量
                if (mid - query(mid) >= k + 1) r = mid;
                else l = mid + 1;
            }
            res[r - 1] = p;
            add(r, 1);
        }
        return res;
    }
}