BIT-Reverse Pairs

2019-12-17 11:07:02

问题描述：

问题求解：

本题可以看作是逆序数问题的强化版本，需要注意的是num[i] > 2 * num[j]，这里有0和负数的情况。

    public int reversePairs(int[] nums) {
        int res = 0;
        int n = nums.length;
        int[] nums_copy = Arrays.copyOf(nums, n);
        TreeMap<Integer, Integer> map = new TreeMap<>();
        Arrays.sort(nums_copy);
        int rank = 0;
        for (int i = 0; i < n; i++) {
            if (i == 0 || nums_copy[i] != nums_copy[i - 1])
                map.put(nums_copy[i], ++rank);
        }
        int[] bit = new int[map.size() + 1];
        for (int i = n - 1; i >= 0; i--) {
            int num = nums[i] % 2 == 0 ? nums[i] / 2 - 1 : (nums[i] - 1) / 2;
            Integer key = map.floorKey(num);
            if (key != null) res += query(bit, map.get(key));
            update(bit, map.get(nums[i]));
        }
        return res;
    }

    private void update(int[] bit, int idx) {
        for (int i = idx; i < bit.length; i += i & -i) {
            bit[i] += 1;
        }
    }

    private int query(int[] bit, int idx) {
        int res = 0;
        for (int i = idx; i > 0; i -= i & -i) {
            res += bit[i];
        }
        return res;
    }