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315. Count of Smaller Numbers After Self


You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].

Example:

Given nums = [5, 2, 6, 1]

To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.

Return the array [2, 1, 1, 0].

思路:

class Solution
class Node
Node left, right;
int val, sum, dup = 1;
public Node(int v) {
val = v;
}
}

public List<Integer> countSmaller(int[] nums) {
Integer[] ans = new Integer[nums.length];
Node root = null;
for (int i = nums.length - 1; i >= 0; i--) {
root = insert(nums[i], root, ans, i, 0);
}
return Arrays.asList(ans);
}

private Node insert(int num, Node node, Integer[] ans, int i, int preSum) {
if (node == null) {
node = new Node(num);
ans[i] = preSum;
} else if (node.val == num) {
node.dup++;
ans[i] = preSum + node.sum;
} else if (node.val > num) {
node.sum++;
node.left = insert(num, node.left, ans, i, preSum);
} else {
node.right = insert(num, node.right, ans, i, preSum + node.dup + node.sum);
}
return


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